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CBSE Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 9 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Definite Integrals
Quiz 9
Let
I
1
=
∫
2
1
1
√
1
+
x
2
d
x
and
I
2
=
∫
2
1
1
x
d
x
. Then
Report Question
0%
I
1
>
I
2
0%
I
2
>
I
1
0%
I
1
=
I
2
0%
I
1
>
2
I
2
Explanation
I
1
=
∫
2
1
1
√
1
+
x
2
d
x
=
(
log
(
x
+
√
1
+
x
2
)
)
2
1
=
log
(
2
+
√
5
)
−
log
(
1
+
√
2
)
=
log
(
(
2
+
√
5
)
(
1
+
√
2
)
)
I
2
=
∫
2
1
1
x
d
x
=
(
log
x
)
2
1
=
log
2
−
log
1
=
log
2
∴
I
2
>
I
1
The value (s) of
∫
1
0
x
4
(
1
−
x
)
4
1
+
x
2
d
x
is (are)
Report Question
0%
22
7
−
π
0%
2
105
0%
0
0%
71
15
−
3
π
2
Explanation
Let
I
=
∫
1
0
x
4
(
1
−
x
)
4
1
+
x
2
d
x
=
∫
1
0
(
x
4
−
1
)
(
1
−
x
)
4
+
(
1
−
x
)
4
(
1
+
x
2
)
d
x
=
∫
1
0
(
x
2
−
1
)
(
1
−
x
)
4
d
x
+
∫
1
0
(
1
+
x
2
−
2
x
)
2
(
1
+
x
2
)
d
x
=
∫
1
0
{
(
x
2
−
1
)
(
1
−
x
)
4
+
(
1
+
x
2
)
−
4
x
+
4
x
2
(
1
+
x
2
)
}
d
x
=
∫
1
0
(
(
x
2
−
1
)
(
1
−
x
)
4
+
(
1
+
x
2
)
−
4
x
+
4
4
1
−
x
2
)
d
x
=
∫
1
0
(
x
6
−
4
x
5
+
5
x
4
−
4
x
2
+
4
−
4
1
+
x
2
)
d
x
=
[
x
7
7
−
4
x
6
6
+
5
x
5
5
−
4
x
3
3
+
4
x
−
4
tan
−
1
x
]
1
0
=
1
7
−
4
6
+
5
5
−
4
3
+
4
−
4
(
π
4
−
0
)
=
22
7
−
π
If
I
=
∫
1
0
x
d
x
8
+
x
3
then the smallest interval in which
I
lies is
Report Question
0%
(
0
,
1
8
)
0%
(
0
,
1
9
)
0%
(
0
,
1
10
)
0%
(
0
,
1
7
)
Explanation
I
=
∫
1
0
x
8
+
x
3
d
x
=
∫
1
0
x
(
x
+
2
)
(
x
2
−
2
x
+
4
)
d
x
=
∫
1
0
(
x
+
2
6
(
x
2
−
2
x
+
4
)
−
1
6
(
x
+
2
)
)
d
x
=
1
6
∫
1
0
(
2
x
−
2
2
(
x
2
−
2
x
+
4
)
+
3
6
x
2
−
2
x
+
4
)
d
x
−
1
6
∫
1
x
+
2
d
x
=
1
2
∫
1
0
2
x
−
2
x
2
−
2
x
+
4
d
x
+
1
2
∫
1
0
1
x
2
−
2
x
+
4
−
1
6
∫
1
0
1
x
+
2
d
x
=
[
1
12
log
(
x
2
−
2
x
+
4
)
−
1
6
log
(
x
+
2
)
]
+
tan
−
1
(
x
−
1
√
3
)
2
√
3
=
1
12
log
(
3
)
−
1
6
log
(
3
)
+
tan
−
1
(
2
√
3
)
2
√
3
−
1
12
log
(
4
)
−
1
6
log
(
2
)
+
tan
−
1
(
−
1
√
3
)
2
√
3
And this lies between
(
0
,
1
9
)
If
I
n
=
∫
π
4
0
tan
n
x
d
x
then
1
I
2
+
I
4
,
1
I
3
+
I
5
,
1
I
4
+
I
6
are in?
Report Question
0%
A
.
P
0%
H
.
P
0%
G
.
P
0%
None of these
Explanation
I
n
=
∫
π
/
4
0
tan
n
−
2
x
tan
2
x
d
x
=
∫
π
/
4
0
tan
n
−
2
x
(
sec
2
x
−
1
)
d
x
=
∫
π
/
4
0
tan
n
−
2
x
sec
2
x
d
x
−
∫
π
/
4
0
tan
n
−
2
x
d
x
I
n
=
[
tan
n
−
1
x
n
−
1
]
π
/
4
0
−
I
n
−
2
∴
I
n
+
I
n
−
2
=
1
n
−
1
Substitute
n
=
4
,
5
,
6
,
.
.
.
.
we get
I
4
+
I
2
,
I
5
+
I
3
,
I
6
+
I
4
,
.
.
.
.
are respectively
1
3
,
1
4
,
1
5
⋯
which are in H.P. and hence their reciprocals are in A.P.
If
I
t
=
∫
π
2
0
sin
2
t
x
sin
2
x
d
x
then ,
I
1
,
I
2
,
I
3
are in
Report Question
0%
A.P.
0%
H.P.
0%
G.P.
0%
None of these
Explanation
Consider
I
t
+
2
+
I
t
−
2
I
t
+
1
∫
π
/
2
0
(
sin
2
t
x
+
sin
2
(
t
+
2
)
x
−
2
sin
2
(
t
+
1
)
x
)
d
x
sin
2
x
∫
π
/
2
0
(
sin
2
(
t
+
2
)
x
−
sin
2
(
t
+
1
)
x
−
sin
2
(
t
+
1
)
x
+
sin
2
t
x
)
d
x
sin
2
x
∫
π
/
2
0
[
sin
(
2
t
+
3
)
x
−
sin
(
2
t
+
1
)
x
]
d
x
sin
x
=
2
∫
π
/
2
0
cos
(
2
t
+
2
)
x
d
x
=
2
(
sin
(
2
t
+
2
)
x
2
(
t
+
1
)
)
π
/
2
0
=
1
(
t
+
1
)
(
0
)
=
0
∴
I
t
,
I
t
+
1
,
I
t
+
2
ϵ
A
.
P
.
Short Cut Method :
Substitute
t
=
1
,
2
,
3
in given integral we get
I
1
=
π
2
,
I
2
=
π
,
I
3
=
3
π
2
I
1
,
I
2
,
I
3
ϵ
A
.
P
.
The value of
∫
π
/
2
0
sin
θ
log
(
sin
θ
)
d
θ
equals
Report Question
0%
log
e
(
1
e
)
0%
log
2
e
0%
log
e
2
−
1
0%
log
e
(
e
2
)
Explanation
Let
I
=
∫
π
2
0
sin
θ
log
(
sin
θ
)
d
θ
=
1
2
∫
π
2
0
sin
θ
log
(
sin
2
θ
)
d
θ
(
∵
log
e
a
m
=
m
log
e
a
)
=
1
2
∫
π
2
0
sin
θ
log
(
1
−
cos
2
θ
)
d
θ
Take
t
=
cos
θ
⇒
sin
θ
d
θ
=
−
d
t
θ
=
0
,
t
=
1
and
θ
=
π
2
,
t
=
0
Then,
I
=
1
2
∫
1
0
log
(
1
−
t
2
)
d
t
=
−
1
2
[
∫
1
0
(
t
2
+
t
4
2
+
t
6
3
+
.
.
.
∞
)
d
t
]
=
−
1
2
[
1
3
+
1
2
⋅
5
+
1
7
⋅
3
+
.
.
.
∞
]
=
−
[
1
2.3
+
1
4
⋅
5
+
1
6
⋅
7
+
.
.
.
.
∞
]
=
−
[
(
1
2
−
1
3
)
+
(
1
4
−
1
5
)
+
(
1
6
−
1
7
)
+
.
.
.
.
∞
]
=
−
1
2
+
1
3
−
1
4
+
1
5
−
1
6
+
1
7
+
.
.
.
.
∞
=
(
1
−
1
2
+
1
3
−
1
4
+
1
5
−
1
6
+
1
7
+
.
.
.
.
∞
)
−
1
=
log
e
2
−
1
=
log
e
2
−
log
e
=
log
e
(
2
e
)
Ans: C
If
f
(
x
)
=
x
∫
1
tan
−
1
t
t
d
t
∀
∈
R
, then the value of
f
(
e
2
)
−
f
(
1
e
2
)
is
Report Question
0%
0
0%
π
2
0%
π
0%
2
π
The value(s) of
∫
1
0
x
4
(
1
−
x
)
4
1
+
x
2
d
x
is (are)
Report Question
0%
22
7
−
π
0%
2
105
0%
0
0%
71
15
−
3
π
2
If
2
f
(
x
)
+
f
(
−
x
)
=
1
x
sin
(
x
−
1
x
)
, then the value of
∫
e
1
e
f
(
x
)
d
x
, is
Report Question
0%
1
0%
0
0%
e
0%
−
1
Explanation
Since,
2
f
(
x
)
+
f
(
−
x
)
=
1
x
sin
(
x
−
1
x
)
...(i)
∴
2
f
(
−
x
)
+
f
(
x
)
=
1
x
sin
(
x
−
1
x
)
(replace
x
by
−
x
) ...(ii)
⟹
f
(
x
)
=
f
(
−
x
)
[subtracting equations (i) and (ii)]
∴
3
f
(
x
)
=
1
x
sin
(
x
−
1
x
)
Hence,
I
=
∫
e
1
e
f
(
x
)
d
x
=
1
3
∫
e
1
e
1
x
sin
(
x
−
1
x
)
d
x
Now, put
x
=
1
t
⇒
d
x
=
−
1
t
2
d
t
∴
I
=
1
3
∫
1
e
e
t
sin
(
1
t
−
t
)
.
(
−
1
t
2
)
d
t
=
1
3
∫
1
e
e
1
t
sin
(
t
−
1
t
)
d
t
=
−
1
3
∫
e
1
e
1
t
sin
(
t
−
1
t
)
d
t
⇒
I
=
−
I
⇒
2
I
=
0
⇒
I
=
0
∴
∫
e
1
e
f
(
x
)
d
x
=
0
.
If
I
=
∫
π
1
/
π
1
x
⋅
sin
(
x
−
1
x
)
d
x
then I is equal to
Report Question
0%
0
0%
π
0%
π
−
1
π
0%
π
+
1
π
Explanation
Let
I
=
∫
π
1
π
1
x
⋅
sin
(
x
−
1
x
)
d
x
Substitute
x
=
1
t
∴
I
=
∫
1
π
π
t
sin
(
1
t
−
t
)
(
−
1
t
2
)
d
t
=
−
∫
π
1
π
1
t
sin
(
t
−
1
t
)
d
t
−
I
⇒
2
I
=
0
⇒
I
=
0
If
x
satisfies the equation
(
∫
1
0
d
t
t
2
+
2
t
cos
α
+
1
)
x
2
−
(
∫
3
−
3
t
2
sin
2
t
t
2
+
1
d
t
)
x
−
2
=
0
for
(
0
<
α
<
π
)
then the value of
x
is?
Report Question
0%
±
√
α
2
sin
α
0%
±
√
2
sin
α
α
0%
±
√
α
sin
α
0%
±
2
√
sin
α
α
Explanation
(
∫
1
0
d
t
t
2
+
2
t
cos
α
+
1
)
x
2
−
(
∫
3
−
3
t
2
sin
2
t
t
2
+
1
d
t
)
x
−
2
=
0
Here,
∫
3
−
3
t
2
sin
2
t
t
2
+
1
d
t
=
0
(
∵
f
(
t
)
=
t
2
sin
2
t
t
2
+
1
is odd
)
So, the equation becomes
(
∫
1
0
d
t
t
2
+
2
t
cos
α
+
1
)
x
2
−
2
=
0
....(1)
Now,
∫
1
0
d
t
t
2
+
2
t
cos
α
+
1
=
∫
1
0
d
t
(
t
+
cos
α
)
2
+
sin
2
α
=
1
sin
α
[
tan
−
1
(
t
+
cos
α
sin
α
)
]
1
0
=
1
sin
α
[
tan
−
1
cot
α
2
−
tan
−
1
cot
α
]
=
1
sin
α
[
π
2
−
α
2
−
π
2
+
α
]
=
α
2
sin
α
So, equation (1) becomes
α
2
sin
α
x
2
−
2
=
0
⇒
x
=
±
2
√
sin
α
α
Hence, option D.
The tangent to the graph of the function
y
=
f
(
x
)
at the point with abscissa x = a forms with the x-axis an angle of
π
/
3
and at the point with abscissa x = b at an angle of
π
/
4
, then the value of the integral,
∫
b
a
f
′
(
x
)
.
f
″
is equal to
Report Question
0%
1
0%
0
0%
\displaystyle -\sqrt 3
0%
-1
Explanation
Given , at
x=a , \frac{dy}{dx}=\tan {\frac{\pi}{3}}=\sqrt{3}
\Rightarrow f'(a)=\sqrt{3}
Also, at
x=b, \frac{dy}{dx}=\tan {\frac{\pi}{4}}=1
\Rightarrow f'(b)=1
Now,
\displaystyle \int_{a}^{b} f'(x).f''(x)dx
Put
f'(x)=t \Rightarrow f''(x)dx=dt
=\displaystyle \int_{\sqrt{3}}^{1} t dt
=-\displaystyle \int_{1}^{\sqrt{3}} t dt
\displaystyle =-[\frac{t^{2}}{2}]_{1}^{\sqrt{3}}
=-1
Let
\displaystyle F\left ( x \right )=f\left ( x \right )+f\left ( \frac{1}{x} \right )
where
\displaystyle f\left ( x \right )=\int_{1}^{x}\frac{\log t}{1+t}dt
Then
F(e)
is equal to?
Report Question
0%
1
0%
2
0%
1/2
0%
0
Explanation
\displaystyle F\left ( x \right )=\int_{1}^{x}\frac{\ln t}{1+t}dt+\int_{1}^{1/x}\frac{\ln t}{1+t}dt
\displaystyle F\left ( x \right )=\int_{1}^{x}\left ( \frac{\ln t}{1+t}+\frac{\ln t}{\left ( 1+t \right )t} \right )dt=\int_{1}^{t}\frac{\ln t}{t}dt=\frac{1}{2}\left ( \ln x \right )^{2}
F\left ( e \right )=1/2
The value of
\displaystyle \int _{ 0 }^{ \pi /2 }{ \frac { d\theta }{ 5+3\cos { \theta } } }
is?
Report Question
0%
\displaystyle \tan ^{ -1 }{ \frac { 1 }{ 2 } }
0%
\displaystyle \tan ^{ -1 }{ \frac { 1 }{ 3 } }
0%
\displaystyle \frac{1}{2}\tan ^{ -1 }{ \frac { 1 }{ 2 } }
0%
\displaystyle \frac{1}{3}\tan ^{ -1 }{ \frac { 1 }{ 3 } }
Explanation
Let
I=\displaystyle \int _{ 0 }^{ \pi /2 }{ \frac { d\theta }{ 5+3\cos { \theta } } } =\int _{ 0 }^{ \pi /2 }{ \frac { d\theta }{ 5+3.\frac { 1-\tan ^{ 2 }{ \frac { \theta }{ 2 } } }{ 1+\tan ^{ 2 }{ \frac { \theta }{ 2 } } } } }
\displaystyle =\int _{ 0 }^{ \pi /2 }{ \frac { \sec ^{ 2 }{ \frac { \theta }{ 2d\theta } } }{ 8+2\tan ^{ 2 }{ \frac { \theta }{ 2 } } } }
\displaystyle =\int _{ 0 }^{ \pi /2 }{ \frac { \frac { 1 }{ 2 } \sec ^{ 2 }{ \left( \frac { \theta }{ 2 } \right) } d\theta }{ 4+\tan ^{ 2 }{ \left( \frac { \theta }{ 2 } \right) } } } =\int _{ 0 }^{ 1 }{ \frac { dt }{ 4+{ t }^{ 2 } } }
[
Put
\displaystyle \tan { \frac { \theta }{ 2 } } =t
So,
\dfrac { 1 }{ 2 } \sec ^{ 2 }{ \dfrac { \theta }{ 2 } } d\theta =dt]
\therefore I=\displaystyle =\frac { 1 }{ 2 } \left[ \tan ^{ -1 }{ \frac { t }{ 2 } } \right] _{ 0 }^{ 1 }
\displaystyle =\frac { 1 }{ 2 } \left( \tan ^{ -1 }{ \frac { 1 }{ 2 } } -\tan ^{ -1 }{ 0 } \right)
= \displaystyle \frac { 1 }{ 2 } \tan ^{ -1 }{ \dfrac { 1 }{ 2 } }
Evaluate :
\displaystyle \underset{0}{\overset{\infty}{\int}} \dfrac{dx}{(1 + x^2)^4}
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0%
\dfrac{\pi}{32}
0%
\dfrac{3 \pi}{32}
0%
\dfrac{5 \pi}{32}
0%
\dfrac{7 \pi}{32}
Let
f:R\rightarrow R^{+}
and
I_{I}=\int^{k}_{1-k}\,xf(x(1-x))\,dx,I_2=\int^{k}_{1-k}f(x(1-x))\,dx
where
2k-1>0
. Then
\dfrac{I_I}{I_2}
is
Report Question
0%
2
0%
k
0%
1/2
0%
1
The value of
\int _{ 0 }^{ \infty }{ \cfrac { \log { x } }{ { a }^{ 2 }+{ x }^{ 2 } } }
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0%
\cfrac{2\pi\log{a}}{a}
0%
\cfrac{\pi\log{a}}{2a}
0%
\pi \log{a}
0%
0
\displaystyle \int_{-1}^{1} \cot^{-1} \left(\dfrac{x+x^{3}}{1+x^{4}}\right)dx
is equal to
Report Question
0%
2\pi
0%
\dfrac{\pi}{2}
0%
0
0%
\pi
\displaystyle \int_{0}^{\pi /2}\sin x\log \left ( \sin x \right )dx=
Report Question
0%
\displaystyle \log _{e}e
0%
\displaystyle \log _{e}2
0%
\displaystyle \log _{e}\left ( e/2 \right )
0%
\displaystyle \log _{e}\left ( 2/e \right )
Explanation
\displaystyle I= \int_{0}^{\pi /2}\sin x\log \left ( \sin x \right )dx
\displaystyle = \frac{1}{2}\int_{0}^{\pi /2}\sin x\log \sin ^{2}x dx
\displaystyle = \frac{1}{2}\int_{0}^{\pi /2}\sin x\log \left ( 1-\cos ^{2}x \right )dx
\displaystyle = \frac{1}{2}\int_{0}^{1}\log \left ( 1-t^{2} \right )dt
where
\displaystyle t= \cos x
\displaystyle = \frac{1}{2}\int_{0}^{1}\left [ -t^{2}-\frac{\left ( t^{2} \right )^{2}}{2}-\frac{\left ( t^{2} \right )^{3}}{3}-\cdots \right ]dt
\displaystyle = -\left [ \frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\cdots \right ]
\displaystyle = -\left [ \left ( \frac{1}{2}-\frac{1}{3} \right )+\left ( \frac{1}{4}-\frac{1}{5} \right )+\left ( \frac{1}{6}-\frac{1}{7} \right )+\cdots \right ]
\displaystyle = -1+\left ( 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots \right )
.
\displaystyle = -\log _{e}e+\log _{e}\left ( 1+1 \right )= \log _{e}\left ( 2/e \right )
\displaystyle \int_{0}^{\infty}\frac{1}{1+x^{n}}dx,\:\forall\:n\:> 1
is equal to?
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0%
\displaystyle2 \int_{0}^{\infty}\frac{1}{1+x}dx
0%
\displaystyle \int_{-\infty}^{\infty}\frac{1}{1+x^{n}}dx
0%
\displaystyle \int_{1}^{\infty}\frac{dx}{(x^{n}-1)^{1/n}}
0%
\displaystyle \int_{0}^{1}\frac{1}{(1-x^{n})^{1/n}}
Explanation
Let
\displaystyle I=\int _{ 0 }^{ \infty } \frac { 1 }{ 1+x^{ n } } dx=\int _{ \infty }^{ 0 } \frac { -dx }{ 1+x^{ n } }
Substitute
\displaystyle x=\frac { 1 }{ y } \Rightarrow dx=-\frac { 1 }{ { y }^{ 2 } }
\displaystyle I=\int _{ 0 }^{ \infty } \frac { 1 }{ y^{ 2 } } \frac { dy }{ 1+\left( \frac { 1 }{ y } \right) ^{ n } } =\int _{ 0 }^{ \infty } \frac { y^{ n-1 } }{ y(1+y^{ n }) } dy
Substitute
t^{ n }-1+y^{ n }\Rightarrow n{ t }^{ n-1 }dt=n{ y }^{ n-1 }dy
\displaystyle I=\int _{ 1 }^{ \infty } \frac { dt }{ t(t^{ n }-t)^{ 1/n } } =\int _{ \infty }^{ 1 } -\frac { 1 }{ t^{ 2 } } .\frac { dt }{ \left( 1-\frac { 1 }{ t^{ n } } \right) ^{ 1/n } }
Substitute
\displaystyle z=\frac { 1 }{ t } \Rightarrow dx=-\frac { 1 }{ { t }^{ 2 } } dt
\displaystyle I=\int _{ 0 }^{ 1 } \frac { dz }{ (1-z^{ n })^{ 1/n } } =\int _{ 0 }^{ 1 } \frac { dx }{ (1-x^{ n })^{ 1/n } }
If
\displaystyle I_1=\int_{0}^{\pi /2}\frac{x}{\sin x}dx
and
\displaystyle I_2=\int_{0}^{\pi /2}\frac{\tan ^{-1}x}{x}dx,
then
\displaystyle \frac{I_{1}}{I_{2}}=
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0%
\displaystyle \frac{1}{2}
0%
1
0%
2
0%
\displaystyle \frac{\pi }{2}
Explanation
Substitute
\displaystyle x=\tan \theta \therefore I_2=\int_{0}^{\pi /4}\frac{\theta }{\tan \theta }.\sec ^{2}\theta d\theta
or
\displaystyle I_{2}=\int_{0}^{\pi /4}\frac{2\theta }{2\sin \theta \cos \theta }d\theta =\int_{0}^{\pi /4} \frac{2\theta }{\sin 2\theta }d\theta
Now substitute
\displaystyle 2\theta =t \therefore I_{2}=\frac{1}{2}\int_{0}^{\pi /2}\frac{t}{\sin t}dt=\frac{1}{2}I_{1}
\displaystyle \therefore \frac{I_{1}}{I_{2}}=2\Rightarrow \left ( C \right )
\displaystyle If \int_{-1}^{1}\frac{g\left ( x \right )}{1+t^{2}}dt= f\left ( x \right ) , where, g\left ( x \right )= \sin x
, then
{f}'\left ( \frac{\pi }{3} \right )
equals
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0%
\displaystyle\frac{\pi }{4}
0%
does not exist
0%
\displaystyle \frac{\pi \sqrt{3}}{4}
0%
None of these
Explanation
\displaystyle f\left( x \right) =\int _{ -1 }^{ 1 }{ \frac { \sin { x } }{ 1+{ t }^{ 2 } } } dt=\sin { x } \int _{ -1 }^{ 1 }{ \frac { 1 }{ 1+{ t }^{ 2 } } } dt
\displaystyle =\sin { x } \left[ \tan ^{ -1 }{ t } \right] _{ -1 }^{ 1 }=\sin { x } \left[ \frac { \pi }{ 4 } +\frac { \pi }{ 4 } \right] =\sin { x } \left[ \frac { \pi }{ 2 } \right]
\displaystyle \therefore f'\left( x \right) =\frac { \pi }{ 2 } \cos { x } \Rightarrow f'\left( \frac { \pi }{ 3 } \right) =\frac { \pi }{ 2 } \frac { 1 }{ 2 } =\frac { \pi }{ 4 }
The value of
\displaystyle \int_{1}^{1/e}f(x)dx+\int_{1}^{e}f(x)dx
where
f(x)
is given as
\displaystyle \frac{log\:x}{1+x}
equals
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0
0%
\displaystyle -\frac{1}{2}
0%
\displaystyle \frac{1}{2}
0%
1
Explanation
Let
\displaystyle l_{ 1 }=\int _{ 1 }^{ 1/e } \frac { logx }{ 1+x } dx
and
\displaystyle l_{ 2 }=\int _{ 1 }^{ 0 } \frac { log\: x }{ 1+x } dx
Now for
\displaystyle l_{ 1 }=\int _{ 1 }^{ 1/e } \frac { log\: t }{ 1+x }
Substitute
\displaystyle x=\frac { 1 }{ t } \Rightarrow dx=-\frac { 1 }{ { t }^{ 2 } }
l_{ 1 }=\int _{ 1 }^{ e } \dfrac { log\: t }{ t(1+t) } dt=\int _{ 1 }^{ e } \dfrac { log\: x }{ x(1+x) } dx
( Replacing
t\rightarrow \: x
)
\displaystyle \therefore l_{ 1 }+l_{ 2 }=\int _{ 1 }^{ e } \frac { log\: x }{ x(1+x) } dx+\int _{ 1 }^{ e } \frac { log\: x }{ 1+x } dx
\displaystyle =\int _{ 1 }^{ e } \frac { (log(x))(1+x) }{ x(1+x) } dx=\int _{ 1 }^{ e } \frac { log\: x }{ x } dx
\displaystyle =\frac { 1 }{ 2 } \left[ (log\: x) \right] _{ 1 }^{ e }=\frac { 1 }{ 2 } [log_{ e }e-log_{ e }1]=\frac { 1 }{ 2 }
The value of
\displaystyle \int_{1/e}^{\tan x}\displaystyle \frac{t}{1+t^{2}}\, dt\, +\, \displaystyle \int_{1/e}^{\cot x}\displaystyle \frac{dt}{t\left ( 1+t^{2} \right )}
is
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0%
1/2
0%
1
0%
\pi /4
0%
none of these
Explanation
\displaystyle \int _{ 1/e }^{ \tan x } \frac { t }{ 1+t^{ 2 } } \, dt\, +\, \int _{ 1/e }^{ \cot x } \frac { dt }{ t\left( 1+t^{ 2 } \right) }
\displaystyle =\int _{ 1/e }^{ \tan x } \frac { t }{ 1+t^{ 2 } } \, dt\, +\, \int _{ 1/e }^{ \cot x } \left( \frac { 1 }{ t } -\frac { 1 }{ 1+{ t }^{ 2 } } \right) dt
=\dfrac { 1 }{ 2 } { \left[ \log { \left( 1+{ t }^{ 2 } \right) } \right] }_{ 1/e }^{ \tan { x } }+{ \left[ \log { t } \right] }_{ 1/e }^{ \cot { x } }-{ \left[ \tan ^{ -1 }{ t } \right] }_{ 1/e }^{ \cot { x } } =1
Evaluate :
\displaystyle \int_{-\frac{1}{\sqrt2}}^{\frac{1}{\sqrt2}}\frac{x^{8}}{1-x^{4}}\times \left [ \sin ^{-1}\left ( 1-2x^{2} \right ) +\cos ^{-1}\left ( 2x\sqrt{1-x^{2}} \right )\right ]dx
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\displaystyle \pi \left [ \frac{1}{2}\log \frac{\sqrt{2}+1}{\sqrt{2}-1}+\tan ^{-1} \frac{1}{\sqrt{2}}-\frac{21}{10\sqrt{2}}\right ]
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\displaystyle \pi \left [ \frac{1}{2}\log \frac{\sqrt{2}+1}{\sqrt{2}-1}+\tan ^{-1} \frac{1}{\sqrt{2}}+\frac{21}{10\sqrt{2}}\right ]
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\displaystyle \pi \left [ \frac{1}{2}\log \frac{\sqrt{2}-1}{\sqrt{2}-1}+\tan ^{-1} \frac{1}{\sqrt{2}}-\frac{21}{10\sqrt{2}}\right ]
0%
\displaystyle \pi \left [ \frac{1}{2}\log \frac{\sqrt{2}-1}{\sqrt{2}-1}+\tan ^{-1} \frac{1}{\sqrt{2}}+\frac{21}{10\sqrt{2}}\right ]
Explanation
In
\displaystyle N^{r}
put
\displaystyle x=\sin \theta ,
then
\displaystyle \sin ^{-1}\left ( 1-2\sin ^{2}\theta \right )+\cos ^{-1}\left ( 2\sin \theta \cos \theta \right )=\sin ^{-1}\left ( \cos 2\theta \right )+\cos ^{-1}\left ( \sin 2\theta \right )
\displaystyle =\frac{\pi }{2}-\cos ^{-1}\left ( \cos 2\theta \right )+\frac{\pi }{2}-\sin ^{-1}\left ( \sin 2\theta \right )
\displaystyle =\pi -2\theta -2\theta =\pi -4\sin ^{-1}x
\displaystyle \therefore I=\int_{-\frac{1}{\sqrt2}}^{\frac{1}{\sqrt2}}\frac{x^{8}}{1-x^{4}}\left [ \pi -4\sin -\mid x \right ]dx
\displaystyle =2 \int_{0}^{\frac{1}{\sqrt{2}}}\pi \frac{x^{8}}{1-x^{4}}dx+0
by Prop.V
\displaystyle =2\pi \int_{0}^{\frac{1}{\sqrt{2}}}\frac{x^{8}-1+1}{1-x^{4}}dx
\displaystyle =2\pi \int_{0}^{\frac{1}{\sqrt{2}}}\left [ -\left ( x^{4}+1 \right )+\frac{1}{\left ( 1-x^{2} \right )\left ( 1+x^{2} \right )} \right ]dx
\displaystyle =2\pi \int_{0}^{\frac{1}{\sqrt{2}}}-\left ( x^{4}+1 \right )+\frac{1}{2}\left \{ \frac{1}{1-x^{2}}+\frac{1}{1+x^{2}} \right \}dx
\displaystyle =2x\left [ -\left ( \frac{x^{5}}{5}+x \right )+\frac{1}{4}\log \frac{1+x}{1-x} +\frac{1}{2}\tan -1x\right ]^{1/\sqrt{2}}_{0}
\displaystyle =\pi \left [ \frac{1}{2}\log \frac{\sqrt{2}+1}{\sqrt{2}-1}+\tan ^{-1} \frac{1}{\sqrt{2}}-\frac{21}{10\sqrt{2}}\right ]
Ans: A
Solve :-
\int_0^1 {\frac{{dx}}{{{{({x^2} + 1)}^{3/2}}}} = }
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0%
1
0%
\frac{1}{{\sqrt 2 }}
0%
\frac{1}{2}
0%
\sqrt 2
Solve:-
\int\limits_0^1 {\frac{{dx}}{{{{({x^2} + 1)}^{3/2}}}}}
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0%
1
0%
\sqrt 2
0%
1/2
0%
\dfrac{1}{\sqrt 2}
If
I= \displaystyle \int_{0}^{1/\sqrt{3}}\displaystyle \frac{dx}{\left ( 1+x^{2} \right )\sqrt{1-x^{2}}}
then
I
is equal to
Report Question
0%
\pi /2
0%
\pi /2\sqrt{2}
0%
\pi /4\sqrt{2}
0%
\pi /4
Explanation
Let
\displaystyle I=\int { \frac { 1 }{ \left( 1+{ x }^{ 2 } \right) \sqrt { 1-{ x }^{ 2 } } } dx }
Substitute
x=\sin { t } \Rightarrow dx=\cos { t } dt
\displaystyle I=\int { \frac { 1 }{ \sin ^{ 2 }{ t } +1 } dt }
Multiply numerator and denominator by
\csc ^{ 2 }{ t }
\displaystyle I=\int { \frac { \csc ^{ 2 }{ t } }{ \csc ^{ 2 }{ t } +1 } dt } =\int { \frac { \csc ^{ 2 }{ t } }{ \cot ^{ 2 }{ t } +2 } dt }
Substitute
u=\cot { t } \Rightarrow du=-\csc ^{ 2 }{ t } dt
\displaystyle I=-\int { \frac { 1 }{ { u }^{ 2 }+2 } du } =-\frac { 1 }{ \sqrt { 2 } } \tan ^{ -1 }{ \frac { u }{ \sqrt { 2 } } }
\displaystyle =-\frac { 1 }{ \sqrt { 2 } } \tan ^{ -1 }{ \frac { \cot { t } }{ \sqrt { 2 } } } =-\frac { 1 }{ \sqrt { 2 } } \tan ^{ -1 }{ \frac { \cot { \sin ^{ -1 }{ x } } }{ \sqrt { 2 } } }
Therefore
\displaystyle \int _{ 0 }^{ 1/\sqrt { 3 } }{ \frac { 1 }{ \left( 1+{ x }^{ 2 } \right) \sqrt { 1-{ x }^{ 2 } } } dx } =\frac { \pi }{ 4\sqrt { 2 } }
\displaystyle \int_{0}^{\pi /3}\frac{\cos \theta }{3+4\sin \theta }d\theta =\lambda \log \frac{3+2\sqrt{3}}{3}
then
\displaystyle \lambda
equals
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0%
\dfrac{1}{2}
0%
\dfrac{1}{3}
0%
\dfrac{1}{4}
0%
\dfrac{1}{8}
Explanation
Given
\displaystyle \int_{0}^{\pi /3}\frac{\cos \theta }{3+4\sin \theta }d\theta =\lambda \log \frac{3+2\sqrt{3}}{3}
....(1)
Consider,
I= \int _{ 0 }^{ \pi /3 } \dfrac { \cos \theta }{ 3+4\sin \theta } d\theta
Substitute
\sin \theta =t
\Rightarrow \cos \theta d\theta =dt
I=\displaystyle \int _{ 0 }^{ \sqrt { 3 } /2 } \frac { dt }{ 3+4t }
Substitute
3+4t=u
4dt=du
\therefore I=\displaystyle \frac {1}{4}\int _{ 3 }^{ 3+2\sqrt { 3 } } \frac { du }{ u }
\displaystyle =\frac{1}{4}[\log { u } ]_{ 3 }^{3+ 2\sqrt { 3 } }
\displaystyle I=\frac{1}{4} [\log { (3+2\sqrt { 3 } )-\log { 3 } }]
\Rightarrow I=\displaystyle \frac{1}{4}\log \frac { 3+2\sqrt { 3 } }{ 3 }
So, on comparing with (1), we get
\lambda =\displaystyle \frac{1}{4}
The value of
\displaystyle \int ^{\tan x}_{1/e}\displaystyle \frac{t\, dt}{1+t^{2}}+\displaystyle \int ^{\cot x}_{1/e}\displaystyle \frac{dt}{t\left ( 1+t^{2} \right )}
is
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0%
\displaystyle \frac{1}{2+tan^{2}x}
0%
1
0%
\pi /4
0%
\displaystyle \frac{2}{\pi }\displaystyle \int ^{1}_{-1}\displaystyle \frac{dt}{1+t^{2}}
Explanation
\displaystyle \int _{ 1/e }^{ \tan x } \frac { t }{ 1+t^{ 2 } } \, dt\, +\, \int _{ 1/e }^{ \cot x } \frac { dt }{ t\left( 1+t^{ 2 } \right) }
\displaystyle =\int _{ 1/e }^{ \tan x } \frac { t }{ 1+t^{ 2 } } \, dt\, +\, \int _{ 1/e }^{ \cot x } \left( \frac { 1 }{ t } -\frac { 1 }{ 1+{ t }^{ 2 } } \right) dt
=\frac { 1 }{ 2 } { \left[ \log { \left( 1+{ t }^{ 2 } \right) } \right] }_{ 1/e }^{ \tan { x } }+{ \left[ \log { t } \right] }_{ 1/e }^{ \cot { x } }-{ \left[ \tan ^{ -1 }{ t } \right] }_{ 1/e }^{ \cot { x } }\\ =1
If
\displaystyle I_{n}= \int_{0}^{\pi /4}\tan ^{n}x\times \sec ^{2}x dx,
then
\displaystyle I_{1}, I_{2}, I_{3},
.......are in
Report Question
0%
A.P.
0%
G.P.
0%
H.P.
0%
none
Explanation
I_{ n }=\int _{ 0 }^{ \pi /4 } \tan ^{ n }{ x } \sec ^{ 2 } xdx
Let,
t=\tan { x }
\Rightarrow \sec ^{ 2 } xdx=dt
Therefore,
I_{ n }=\int _{ 0 }^{ 1 }{ { t }^{ n } } dt={ \left( \dfrac { t }{ n+1 } \right) }_{ 0 }^{ 1 }=\frac { 1 }{ n }
Therefore,
I_{ 1 },I_{ 2 },I_{ 3 }\cdots
are
\dfrac { 1 }{ 2 } ,\dfrac { 1 }{ 3 } ,\dfrac { 1 }{ 4 } ,\cdots
which are in H.P
Ans: C
The value of the integral
\displaystyle \int_{\alpha }^{\beta }\displaystyle \dfrac{dx}{\sqrt{\left ( x-\alpha \right )\left ( \beta -x \right )}}
for
\beta > \alpha
, is
Report Question
0%
\sin ^{-1}\: \alpha /\beta
0%
\pi /2
0%
\sin ^{-1}\beta /2\alpha
0%
\pi
Explanation
Given :
\displaystyle \int _{ \alpha }^{ \beta }{ \cfrac { dx }{ \sqrt { (x-\alpha )(\beta -x) } } }
\displaystyle I=\int _{ \alpha }^{ \beta }{ \cfrac { dx }{ \sqrt { (x-\alpha )(\beta -x) } } } =\int _{ \alpha }^{ \beta }{ \cfrac { dx }{ \sqrt { -\alpha \beta +x(\beta +\alpha )-{ x }^{ 2 } } } }
\displaystyle I=\int _{ \alpha }^{ \beta }{ \cfrac { dx }{ \sqrt { \cfrac { { (\alpha +\beta ) }^{ 2 } }{ 4 } -\alpha \beta -\left[ x-\cfrac { \beta +\alpha }{ 2 } \right] ^{ 2 } } } }
\displaystyle I=\int _{ \alpha }^{ \beta }{ \cfrac { dx }{ \sqrt { \cfrac { (\beta -\alpha )^{ 2 } }{ 4 } -\left[ x-\cfrac { (\beta +\alpha ) }{ 2 } \right] ^{ 2 } } } }
we know that,
\displaystyle \int \dfrac{1}{\sqrt{a^2-x^2}}dx =\sin^{-1}\left(\dfrac {x}{a}\right)+c
I=\sin ^{ -1 }{ \left[ \cfrac { x-\left( \cfrac { \beta +\alpha }{ 2 } \right) }{ \left( \cfrac { \beta -\alpha }{ 2 } \right) } \right] } _{ \alpha }^{ \beta }=\sin ^{ -1 }{ \left[ \cfrac { \left( \cfrac { \beta -\alpha }{ 2 } \right) }{ \left( \cfrac { \beta +\alpha }{ 2 } \right) } \right] } -\sin ^{ -1 }{ \left[ \cfrac { \left( \cfrac { \alpha -\beta }{ 2 } \right) }{ \left( \cfrac { \beta -\alpha }{ 2 } \right) } \right] }
I=\sin ^{ -1 }{ (1)-\sin ^{ -1 }{ (-1) } }
I=\cfrac { \pi }{ 2 } +\cfrac { \pi }{ 2 } =\pi
Hence the correct answer is
\pi
\displaystyle \int_{\pi /6}^{\pi /4}\frac{dx}{\sin 2x}
is equal to
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0%
\displaystyle \frac{1}{2}\log \left ( -1 \right )
0%
\displaystyle \log \left ( -1 \right )
0%
\displaystyle \log 3
0%
\displaystyle \frac {1}{2} \log \sqrt{3}
Explanation
Let
\displaystyle I=\int { \frac { 1 }{ \sin { 2x } } dx } =\int { \csc { 2x } dx }
Put
2x=t\Rightarrow 2dx=dt
\displaystyle I=\frac { 1 }{ 2 } \int { \csc { t } dt }
Multiply numerator and denominator by
\cot { t } +\csc { t }
\displaystyle I=\frac { 1 }{ 2 } \int { -\frac { -\csc ^{ 2 }{ t } -\cot { t } \csc { t } }{ \cot { t } +\csc { t } } dt }
Put
\cot { t } +\csc { t } =u\Rightarrow \left( -\csc ^{ 2 }{ t } -\cot { t } \csc { t } \right) dt=du
\displaystyle I=-\frac { 1 }{ 2 } \int { \frac { 1 }{ u } du } =-\frac { 1 }{ 2 } \log { u } =-\frac { 1 }{ 2 } \log { \left( \cot { t } +\csc { t } \right) }
\displaystyle =-\frac { 1 }{ 2 } \log { \left( \cot { 2x } +\csc { 2x } \right) } =-\frac { 1 }{ 2 } \log { \left( \cot { x } \right) }
Hence
\displaystyle \int _{ \pi /6 }^{ \pi /4 }{ \frac { 1 }{ \sin { 2x } } dx } =-\frac { 1 }{ 2 } { \left[ \log { \left( \cot { x } \right) } \right] }_{ \pi /6 }^{ \pi /4 }=\frac { 1 }{ 2 } \log { \sqrt { 3 } }
Value of
\displaystyle \int_{0}^{1}\displaystyle \frac{dx}{\left ( 1+x^{2} \right )\sqrt{1-x^{2}}}
is?
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0%
\dfrac{\pi}{2\sqrt{2}}
0%
\dfrac{\pi}{\sqrt{2}}
0%
\sqrt{2}\pi
0%
2\sqrt{2\pi }
Explanation
Let
\displaystyle I=\int { \frac { dx }{ \left( 1+x^{ 2 } \right) \sqrt { 1-x^{ 2 } } } }
Substitute
\displaystyle x=\frac { 1 }{ t } \Rightarrow dx=-\frac { 1 }{ { t }^{ 2 } } dt
\displaystyle I=\int { \frac { -tdt }{ \left( { t }^{ 2 }+1 \right) \sqrt { { t }^{ 2 }-1 } } }
Substitute
{ t }^{ 2 }-1={ u }^{ 2 }\Rightarrow 2tdt=2udu
\displaystyle I=-\int { \frac { udu }{ \left( { u }^{ 2 }+1 \right) u } } =-\int { \frac { 1 }{ { u }^{ 2 }+{ \left( \sqrt { 2 } \right) }^{ 2 } } du }
\displaystyle =-\frac { 1 }{ \sqrt { 2 } } \tan ^{ -1 }{ \left( \frac { u }{ \sqrt { 2 } } \right) } =-\frac { 1 }{ \sqrt { 2 } } \tan ^{ -1 }{ \left( \frac { \sqrt { 2 } x }{ \sqrt { 1-{ x }^{ 2 } } } \right) }
Therefore
\displaystyle \int _{ 0 }^{ 1 }{ Idx } =\frac { \pi }{ 2\sqrt { 2 } }
If
3+2\displaystyle\int _{ 0 }^{ 1 }{ { x }^{ 2 }{ e }^{ -x^{ 2 } } } dx=\displaystyle\int _{ 0 }^{ 1 }{ { e }^{ -x^{ 2 } } } dx
then the value of
\beta
is
Report Question
0%
\dfrac{1}{e}
0%
e
0%
\dfrac{1}{2e}
0%
2e
If
I_n=\displaystyle\int^1_0\dfrac{dx}{(1+x^2)^n}; n\in N
, then which of the following statements hold good?
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0%
2n I_{n+1}=2^{-n}+(2n-1)I_n
0%
I_2=\dfrac{\pi}{8}+\dfrac{1}{4}
0%
I_2=\dfrac{\pi}{8}-\dfrac{1}{4}
0%
I_3=\dfrac{\pi}{16}-\dfrac{5}{48}
If
\int { \frac { { x }^{ 1/2 } }{ { x }^{ 1/2 }-{ x }^{ 1/3 } } } dx=Ax\frac { 6 }{ 5 } { x }^{ 5/6 }+\frac { 3 }{ 2 } { x }^{ 2/3 }+B\sqrt { x } +{ Cx }^{ 1/3 }+{ Dx }^{ 1/6 }+E
In
\left( { x }^{ 1/6 }-1 \right) =k,
then A+B+C+D+E=
Report Question
0%
6
0%
12
0%
18
0%
17
The value of
\displaystyle \int_{1/2}^{1}\displaystyle \frac{dx}{x\sqrt{3x^{2}+2x-1}}
is?
Report Question
0%
\pi /2
0%
\pi /3
0%
\pi /6
0%
\pi /\sqrt{2}
Explanation
Let
\displaystyle I=\int { \frac { 1 }{ x\sqrt { { 3x }^{ 2 }+2x-1 } } } dx=\int { \frac { 1 }{ x\sqrt { { \left( \sqrt { 3 } x+\frac { 1 }{ \sqrt { 3 } } \right) }^{ 2 }-\frac { 4 }{ 3 } } } } dx
Put
\displaystyle t=\sqrt { 3 } x+\frac { 1 }{ \sqrt { 3 } } \Rightarrow dt=\sqrt { 3 } dt
\displaystyle I=\frac { 1 }{ \sqrt { 3 } } \int { -\frac { 1 }{ \left( \sqrt { 3 } -3t \right) \sqrt { { t }^{ 2 }-\frac { 4 }{ 3 } } } } dt
\displaystyle =-3\int { \frac { 1 }{ \left( \sqrt { 3 } -3t \right) \sqrt { { t }^{ 2 }-\frac { 4 }{ 3 } } } } dt
\displaystyle I=3\int { \frac { 2 }{ 3\sqrt { 3 } { s }^{ 2 }+\sqrt { 3 } } ds } =2\sqrt { 3 } \int { \frac { 1 }{ 3{ s }^{ 2 }+1 } ds }
\displaystyle =2\tan ^{ -1 }{ \left( \sqrt { 3 } s \right) } =\tan ^{ -1 }{ \left( \frac { x-1 }{ \sqrt { 3{ x }^{ 2 }+2x-1 } } \right) }
\displaystyle \therefore \int _{ 1/2 }^{ 1 } { \frac { 1 }{ x\sqrt { { 3x }^{ 2 }+2x-1 } } } { dx } =\frac { \pi }{ 6 }
If
I= \displaystyle \int_{1}^{\infty }\displaystyle \frac{x^{2}-2}{x^{3}\sqrt{x^{2}-1}}\: dx
, then
I
equals
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0%
-1
0%
0
0%
\pi /2
0%
\pi -\sqrt{3}
Explanation
\displaystyle I=\int _{ 1 }^{ \infty }{ \frac { { x }^{ 2 }-2 }{ { x }^{ 3 }\sqrt { { x }^{ 2 }-1 } } dx }
Put Put
x=\sec { u } \Rightarrow dx=\tan { u } \sec { u } du
\displaystyle I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos ^{ 2 }{ u\left( \sec ^{ 2 }{ u } -2 \right) } du } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( 1-\sin ^{ 2 }{ u } \right) \left( \sec ^{ 2 }{ u } -2 \right) du }
\displaystyle =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( 2\sin ^{ 2 }{ u } -\tan ^{ 2 }{ u } +\sec ^{ 2 }{ u } -2 \right) } du=-\int _{ 0 }^{ \frac { \pi }{ 2 } }{ du } +2\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sin ^{ 2 }{ u } du }
\displaystyle =-\int _{ 0 }^{ \frac { \pi }{ 2 } }{ du } -\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos { 2udu } } +\int _{ 0 }^{ \frac { \pi }{ 2 } }{ du } =-{ \left[ \frac { \sin { 2u } }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }=0
If
0< \alpha < \pi /2
then the value of
\displaystyle \int_{0}^{\alpha }\displaystyle \frac{dx}{1-\cos x\cos \alpha }
is
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0%
\pi /\alpha
0%
\pi /2\sin \alpha
0%
\pi /2\cos \alpha
0%
\pi /2\alpha
Explanation
Let
\displaystyle I=\int _{ 0 }^{ \alpha } \dfrac { dx }{ 1-\cos x\cos \alpha }
\displaystyle =\int _{ 0 }^{ \alpha }{ \dfrac { dx }{ \left( \cos ^{ 2 }{ \dfrac { x }{ 2 } } +\sin ^{ 2 }{ \dfrac { x }{ 2 } } \right) -\cos { \alpha } \left( \cos ^{ 2 }{ \dfrac { x }{ 2 } } -\sin ^{ 2 }{ \dfrac { x }{ 2 } } \right) } }
\displaystyle =\int _{ 0 }^{ \alpha }{ \dfrac { dx }{ \left( 1-\cos { \alpha } \right) \cos ^{ 2 }{ \dfrac { x }{ 2 } } +2\cos ^{ 2 }{ \dfrac { \alpha }{ 2 } } \sin ^{ 2 }{ \dfrac { x }{ 2 } } } }
\displaystyle =\dfrac { 1 }{ 2 } \int _{ 0 }^{ \alpha }{ \dfrac { \sec ^{ 2 }{ \dfrac { \alpha }{ 2 } } \sec ^{ 2 }{ \dfrac { x }{ 2 } } }{ \tan ^{ 2 }{ \dfrac { \alpha }{ 2 } } +\tan ^{ 2 }{ \dfrac { x }{ 2 } } } dx }
Substitute
\displaystyle \tan { \dfrac { x }{ 2 } } =t
\displaystyle I=\int _{ 0 }^{ \tan { \dfrac { \alpha }{ 2 } } }{ \dfrac { \sec ^{ 2 }{ \dfrac { \alpha }{ 2 } } }{ \tan ^{ 2 }{ \dfrac { \alpha }{ 2 } } +{ t }^{ 2 } } dt } =\sec ^{ 2 }{ \dfrac { \alpha }{ 2 } } \cot { \dfrac { \alpha }{ 2 } } { \left[ \tan ^{ -1 }{ \dfrac { 1 }{ \tan { \dfrac { \alpha }{ 2 } } } } \right] }_{ 0 }^{ \tan { \dfrac { \alpha }{ 2 } } }
\displaystyle =\dfrac { \pi }{ 2\sin { \alpha } }
Value of
\displaystyle \int_{a}^{\infty }\displaystyle \frac{dx}{x^{4}\sqrt{a^{2}+x^{2}}}
is
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\displaystyle \frac{2+\sqrt{2}}{3a^{4}}
0%
\displaystyle \frac{2-\sqrt{2}}{3a^{2}}
0%
\displaystyle \frac{2-\sqrt{2}}{3a^{4}}
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\displaystyle \frac{\sqrt{2}+1}{3a^{2}}
Explanation
Let
\displaystyle I=\int _{ a }^{ \infty }{ \frac { 1 }{ { x }^{ 4 }\sqrt { { a }^{ 2 }+x^{ 2 } } } } dx
Substitute
x=a\tan { t\Rightarrow } dx=a\sec ^{ 2 }{ t } dt
\displaystyle I=a\int _{ \frac { \pi }{ 4 } }^{ \frac { \pi }{ 2 } }{ \frac { \cot ^{ 3 }{ t } \csc { t } }{ { a }^{ 5 } } } dt
\displaystyle =\frac { 1 }{ { a }^{ 4 } } \int _{ \frac { \pi }{ 4 } }^{ \frac { \pi }{ 2 } }{ \cot { t } \csc { t } \left( \csc ^{ 2 }{ t-1 } \right) dt }
Substitute
\upsilon =\csc { t } \Rightarrow d\upsilon =-\cot { t } \csc { t } dt
\displaystyle I=\frac { 1 }{ { a }^{ 4 } } \int _{ \sqrt { 2 } }^{ 1 }{ \left( { \upsilon }^{ 2 }-1 \right) } d\upsilon
\displaystyle =\frac { 1 }{ { a }^{ 2 } } \left[ \frac { { \upsilon }^{ 2 } }{ 3 } -\upsilon \right] _{ \sqrt { 2 } }^{ 1 }{ =\frac { 2-\sqrt { 2 } }{ { 3a }^{ 4 } } }
The value of
\displaystyle \int_{-4}^{-5}e^{\left ( x+5 \right )^{2}}dx+3\displaystyle \int_{1/3}^{2/3}e^{9\left ( x-2/3 \right )^{2}}dx
is
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0%
2/5
0%
1/5
0%
1/2
0%
none of these
Explanation
Let
I=\int _{ -4 }^{ -5 } e^{ \left( x+5 \right) ^{ 2 } }dx+3\int _{ 1/3 }^{ 2/3 } e^{ 9\left( x-2/3 \right) ^{ 2 } }dx={ I }_{ 1 }+{ I }_{ 2 }
Where
{ I }_{ 1 }=\int _{ -4 }^{ -5 } e^{ \left( x+5 \right) ^{ 2 } }dx
Put
x+5=t
{ I }_{ 1 }=-\int _{ 0 }^{ 1 }{ { e }^{ { t }^{ 2 } } } dt
And
{ I }_{ 2 }=3\int _{ 1/3 }^{ 2/3 } e^{ 9\left( x-2/3 \right) ^{ 2 } }dx
Put
-3x+2=u
{ I }_{ 2 }=\int _{ 0 }^{ 1 }{ { e }^{ { u }^{ 2 } } } du
\therefore I=0
Value of
\displaystyle \int_{0}^{16}\displaystyle \frac{x^{1/4}}{1+x^{1/2}}\: dx
is
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0%
\displaystyle \frac{8}{3}
0%
\displaystyle \frac{4}{3}\tan ^{-1}\: 2
0%
4\displaystyle \left ( \displaystyle \frac{2}{3}+\tan ^{-1}\: 2 \right )
0%
4\displaystyle \left ( \displaystyle \frac{2}{3}-\tan ^{-1}\: 2 \right )
Explanation
\displaystyle I=\int _{ 0 }^{ 16 }{ \frac { \sqrt [ 4 ]{ x } }{ 1+\sqrt { x } } dx }
Put
\displaystyle u=\sqrt [ 4 ]{ x } \Rightarrow du=\frac { 1 }{ 4{ x }^{ \frac { 3 }{ 4 } } } dx
\displaystyle I=4\int _{ 0 }^{ 2 }{ \left( { u }^{ 2 }+\frac { 1 }{ { u }^{ 2 }+1 } -1 \right) } du
\displaystyle =4\int _{ 0 }^{ 2 }{ { u }^{ 2 } } du+4\int _{ 0 }^{ 2 }{ \frac { 1 }{ { u }^{ 2 }+1 } du } -4\int _{ 0 }^{ 2 }{ du }
\displaystyle =4\left[ \frac { { u }^{ 3 } }{ 3 } \right] _{ 0 }^{ 2 }{ + }4\left[ \tan ^{ -1 }{ u } \right] _{ 0 }^{ 2 }-4\left[ u \right] _{ 0 }^{ 2 }
\displaystyle =4\left( \frac { 2 }{ 3 } +\tan ^{ -1 }{ 2 } \right)
The solution of
x
of the equation
\displaystyle \int_{\sqrt{2}}^{x}{\dfrac{dt}{t\sqrt{t^{2}-1}}}=\dfrac{\pi}{2}
is
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0%
2\sqrt{2}
0%
2
0%
\pi
0%
-\sqrt{2}
\int_{0}^{1}\frac{1}{ax+b}dx is
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0%
zero
0%
log_{e}\frac{a+b}{b}
0%
log_{e}(ax+b)
0%
\frac{1}{a}log_{e}\left ( \frac{a+b}{b} \right )
\int _{ 0 }^{ 8 }{ \left[ \sqrt { t } \right] } dt
at equals to (where [.] greatest integer function.)
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0%
28
0%
11
0%
2
0%
8
If
\displaystyle \frac{\pi }{4}< \alpha < \displaystyle \frac{\pi }{2}
, value of
\displaystyle \int_{-\pi /2}^{\pi /2}\displaystyle \frac{\sin 2x}{\sqrt{1+\sin 2\alpha \sin x}}
is
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0%
-\displaystyle \frac{4}{3}\tan \alpha \sec \alpha
0%
-\displaystyle \frac{4}{3}\cot \alpha cosec\alpha
0%
-\displaystyle \frac{4}{3}\tan \alpha cosec\alpha
0%
-\displaystyle \frac{4}{3}\cot \alpha \sec \alpha
Explanation
Let
\displaystyle I=\int _{ -\pi /2 }^{ \pi /2 } \frac { \sin 2x }{ \sqrt { 1+\sin 2\alpha \sin x } } =2\int _{ -\pi /2 }^{ \pi /2 } \frac { \sin { x } \cos { x } }{ \sqrt { 1+\sin 2\alpha \sin x } } dx
Substitute
t=\sin { x } \Rightarrow dt=\cos { x } dx
\displaystyle I=2\int _{ -1 }^{ 1 }{ \frac { t }{ \sqrt { t\sin { 2\alpha } +1 } } dt }
Substitute
u=t\sin { 2\alpha } \Rightarrow du=\sin { 2\alpha } dt
\displaystyle I=\int _{ 1-\sin { 2\alpha } }^{ 1+\sin { 2\alpha } }{ 2\csc ^{ 2 }{ 2\alpha } \frac { u-1 }{ \sqrt { u } } }
\displaystyle { \left[ \frac { 4 }{ 3 } { u }^{ \frac { 3 }{ 2 } }\csc ^{ 2 }{ 2\alpha } -4\sqrt { 5 } \csc ^{ 2 }{ 2\alpha } \right] }_{ 1-\sin { 2\alpha } }^{ 1+\sin { 2\alpha } }
\displaystyle =-\frac { 4 }{ 3 } \cot { \alpha } \csc { \alpha }
If
I_{1}= \displaystyle \int_{x}^{1}\displaystyle \frac{dt}{1+t^{2}}
and
I_{2}= \displaystyle \int_{1}^{1/x}\displaystyle \frac{dt}{1+t^{2}}
for
x> 0
, then
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0%
I_{1}= I_{2}
0%
I_{1}> I_{2}
0%
I_{2}> I_{1}
0%
I_{2}= \left ( \pi /2 \right )-\tan ^{-1}x
If
I_{1}= \displaystyle \int_{0}^{\infty }\displaystyle \frac{dx}{1+x^{4}}
and
I_{2}= \displaystyle \int_{0}^{\infty }\displaystyle \frac{x^{2}}{1+x^{4}}\: dx
, then
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0%
I_{1}= I_{2}
0%
I_{1}=2 I_{2}
0%
2I_{1}= I_{2}
0%
none of these
Explanation
I_{2}=\displaystyle \int _{0}^{\infty} \dfrac{x^2}{1+x^4}d{x}=\int^{\infty}_{0}\dfrac{\dfrac{1}{x^2}}{1+\dfrac{1}{x^4}}d{x}
Put
t=\dfrac{1}{x}\implies d{t}=-\dfrac{d{x}}{x^2}
I_{2}=\displaystyle\int _{\infty}^{0} \dfrac{-d{t}}{1+t^4}=\int_{0}^{\infty} \dfrac{d{x}}{1+x^4}=I_{1}
If
\displaystyle \int_{0}^{\infty }\displaystyle \frac{\log \left ( 1+x^{2} \right )}{1+x^{2}}\: dx= \lambda \displaystyle \int_{0}^{1}\displaystyle \frac{\log \left ( 1+x \right )}{1+x^{2}}\: dx
then
\lambda
equals
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0%
4
0%
\pi
0%
8
0%
2\pi
Explanation
Let
I_1=\displaystyle \int_{0}^{\infty }\displaystyle \frac{\log \left ( 1+x^{2} \right )}{1+x^{2}}\: dx
and
I_2= \displaystyle \int_{0}^{1}\displaystyle \frac{\log \left ( 1+x \right )}{1+x^{2}}\: dx
Put
x=\tan\theta
in both integral
\Rightarrow dx=\sec^2\theta d\theta
I_1=\displaystyle \int_{0}^{\frac{\pi}{2} }\displaystyle \frac{\log \left ( \sec^2\theta \right )}{\sec^2\theta}\cdot \sec^2\theta d\theta=-\int_0^{\frac{\pi}{2}}\log\cos^2\theta d\theta ......(1)
Also using
\left( \int_0^a f(x)dx=\int_0^af(a-x)dx \right)
I_1 =\displaystyle -\int_0^{\frac{\pi}{2}}\log \sin^2\theta d\theta ........(2)
Adding (1) and (2) we get
2I_1 = \displaystyle -2 \int_0^{\frac{\pi}{2}}\log (\sin\theta\cos\theta)d\theta
\Rightarrow\displaystyle I_1=-\int_0^{\frac{\pi}{2}}\log\frac{\sin2\theta}{2} d\theta =\log 2\int_0^{\frac{\pi}{2}}d\theta- \int_0^{\frac{\pi}{2}} \log(\sin2\theta)d\theta
Put
2\theta =x\Rightarrow 2d\theta=dx
\Rightarrow \displaystyle I_1= \frac{\pi}{2}\log 2 -\frac{1}{2}\int_0^{\pi}\log(\sin\theta)d\theta=\frac{\pi}{2}\log 2 -\int_0^{\frac{\pi}{2}}\log(\sin\theta)d\theta=\frac{\pi}{2}\log 2+\frac{I_1}{2}
, using (2)
\therefore \displaystyle I_1 =\pi\log 2
Now
I_2=\displaystyle \int_0^{\frac{\pi}{4}}\log(1+\tan\theta)d\theta
\Rightarrow \displaystyle I_2 =\int_0^{\frac{\pi}{4}}\log(1+\tan(\frac{\pi}{4}-\theta))d\theta=\int_0^{\frac{\pi}{4}}\log\left(1+\frac{1-\tan\theta}{1+\tan\theta}\right)d\theta
\displaystyle \Rightarrow I_2=\int_0^{\frac{\pi}{4}}\log\left(\frac{2}{1+\tan\theta}\right)d\theta=\log2\int_0^{\frac{\pi}{4}}d\theta-\displaystyle \int_0^{\frac{\pi}{4}}\log(1+\tan\theta)d\theta
\Rightarrow \displaystyle I_2=\frac{\pi}{4}\log 2-I_2 \Rightarrow I_2=\frac{\pi}{8}\log 2
Hence
\lambda = \cfrac{I_1}{I_2}=8
0:0:2
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
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