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CBSE Questions for Class 12 Commerce Maths Integrals Quiz 8 - MCQExams.com
CBSE
Class 12 Commerce Maths
Integrals
Quiz 8
∫
5
5
2
√
(
25
−
x
2
)
3
x
4
d
x
is equal to
Report Question
0%
π
6
0%
2
π
3
0%
5
π
6
0%
π
3
Explanation
I
=
∫
5
5
2
√
(
25
−
x
2
)
3
x
4
d
x
Let
x
=
5
sin
θ
∴
d
x
=
5
cos
θ
d
θ
∴
I
=
∫
π
2
π
6
5
3
cos
3
θ
.5
cos
θ
5
4
sin
4
θ
d
θ
=
∫
π
2
π
6
cot
2
θ
(
c
o
s
e
c
2
θ
−
1
)
d
θ
=
∫
π
2
π
6
cot
2
θ
c
o
s
e
c
2
θ
d
θ
−
∫
π
2
π
6
cot
2
θ
d
θ
=
∫
π
2
π
6
cot
2
θ
c
o
s
e
c
2
θ
d
θ
−
∫
π
2
π
6
(
c
o
s
e
c
2
θ
−
1
)
=
[
−
cot
3
θ
3
+
cot
θ
+
θ
]
π
2
π
6
=
−
0
+
0
+
π
2
−
(
−
3
√
3
3
+
√
3
+
π
6
)
=
π
3
The minimum value of the function f(x) =
∫
x
0
d
θ
c
o
s
θ
+
∫
π
/
2
x
d
θ
s
i
n
θ
where
x
∈
[
0
,
π
2
]
,
is
Report Question
0%
2
l
n
(
√
2
+
1
)
0%
l
n
(
2
√
2
+
2
)
0%
l
n
(
√
3
+
2
)
0%
l
n
(
√
2
+
3
)
Explanation
f(x) =
∫
x
0
d
θ
c
o
s
θ
+
∫
π
/
2
x
d
θ
s
i
n
θ
f
′
(
x
)
=
1
c
o
s
x
−
1
s
i
n
x
=
s
i
n
x
−
c
o
s
x
s
i
n
x
c
o
s
x
f
′
(
x
)
change sign
−
t
0
+
a
t
x
=
π
4
so f(x) takes minimum value at x =
π
4
f
(
x
)
=
∫
π
/
4
0
s
e
c
θ
d
θ
+
∫
π
/
2
π
/
4
c
o
s
e
c
θ
d
θ
=
(
l
n
(
s
e
c
θ
+
t
a
n
θ
)
)
π
/
4
0
+
(
l
n
(
c
o
s
e
c
θ
−
c
o
t
θ
)
)
π
/
2
π
/
4
=
l
n
(
√
2
+
1
)
−
l
n
(
1
+
0
)
+
l
n
1
−
l
n
(
√
2
−
1
)
=
l
n
[
√
2
+
1
√
2
−
1
]
=
l
n
(
√
2
+
1
)
2
=
2
l
n
(
√
2
+
1
)
If
∫
1
0
tan
−
1
x
x
d
x
is equal to
Report Question
0%
∫
π
2
0
sin
x
x
d
x
0%
∫
π
2
0
x
sin
x
d
x
0%
1
2
∫
π
2
0
sin
x
x
d
x
0%
1
2
∫
π
2
0
x
sin
x
d
x
Explanation
I
=
∫
1
0
tan
−
1
x
x
d
x
Putting
x
=
tan
θ
or
d
x
=
sec
2
θ
d
θ
, we get
I
=
∫
π
4
0
θ
tan
θ
sec
2
θ
d
θ
=
∫
π
4
0
2
θ
sin
2
θ
d
θ
Putting
2
θ
=
t
,i.e.,
2
d
θ
=
d
t
, we get
I
=
1
2
∫
π
2
0
t
sin
t
d
t
=
1
2
∫
π
2
0
x
sin
x
d
x
The value of the integral
∫
log
5
0
e
x
√
e
x
−
1
e
x
+
3
d
x
is
Report Question
0%
3
+
2
π
0%
4
−
π
0%
2
+
π
0%
none of these
Explanation
Putting
e
x
−
1
=
t
2
in the given integral, we have
∫
log
5
0
e
x
√
e
x
−
1
e
x
+
3
d
x
=
2
∫
2
0
t
2
t
2
+
4
d
t
=
2
(
∫
2
0
1
d
t
−
4
∫
2
0
d
t
t
2
+
4
)
=
2
[
(
t
−
2
tan
−
1
(
t
2
)
)
2
0
]
=
2
[
(
2
−
2
×
π
4
)
]
=
4
−
π
The value of the integral
∫
1
√
3
0
d
x
(
1
+
x
2
)
√
1
−
x
2
Report Question
0%
π
2
√
2
0%
π
4
√
2
0%
π
8
√
2
0%
none of these
Explanation
Orl putting
x
=
sin
θ
,we get
d
x
=
cos
θ
d
θ
Integral (without limits)
=
∫
cos
θ
d
θ
(
1
+
sin
2
θ
)
(
cos
θ
)
=
∫
d
θ
1
+
sin
2
θ
=
∫
c
o
s
e
c
2
θ
d
θ
2
+
cot
2
=
∫
−
d
t
2
+
t
2
where
t
=
cot
θ
=
−
1
√
2
tan
−
1
t
√
2
=
−
1
√
2
tan
−
1
cot
θ
√
2
=
−
1
√
2
tan
−
1
1
√
2
(
√
1
−
x
2
x
)
∴
Definite integral
=
−
1
√
2
tan
−
1
1
+
1
√
2
tan
−
1
∞
=
−
π
4
√
2
+
π
2
√
2
=
π
4
√
2
Let
I
1
=
∫
1
0
e
x
d
x
1
+
x
and
I
2
=
∫
1
0
x
2
d
x
e
x
3
(
2
−
x
3
)
.
Then
I
1
I
2
is equal to?
Report Question
0%
3
e
0%
e
3
0%
3
e
0%
1
3
e
Explanation
In
I
2
=
∫
1
0
x
2
d
x
e
x
3
(
2
−
x
3
)
Substitute
1
−
x
3
=
t
⇒
−
3
x
2
d
x
=
d
t
We get
I
2
=
−
1
3
∫
0
1
d
t
e
1
−
t
(
1
+
t
)
=
1
3
e
∫
1
0
e
t
d
t
1
+
t
=
1
3
e
∫
1
0
e
x
d
x
1
+
x
As
I
1
=
∫
1
0
e
x
d
x
1
+
x
Therefore
I
1
I
2
=
3
e
The value of
∫
∞
0
d
x
1
+
x
4
is
Report Question
0%
same as that of
∫
∞
0
x
2
+
1
d
x
1
+
x
4
0%
π
2
√
2
0%
same as that of
∫
∞
0
x
2
d
x
1
+
x
4
0%
π
√
2
Explanation
I
=
∫
∞
0
d
x
1
+
x
4
....(1)
∫
∞
0
x
2
+
1
−
x
2
1
+
x
4
d
x
=
∫
∞
0
x
2
1
+
x
4
d
x
+
∫
∞
0
1
−
x
2
1
+
x
4
d
x
....(2)
Let
I
1
=
∫
∞
0
x
2
1
+
x
4
d
x
and
I
2
=
∫
∞
0
1
−
x
2
1
+
x
4
d
x
I
2
=
∫
∞
0
1
x
2
−
1
1
x
2
+
x
2
d
x
I
2
=
∫
∞
0
1
x
2
−
1
1
x
2
+
x
2
+
2
−
2
d
x
I
2
=
∫
∞
0
1
x
2
−
1
(
x
+
1
x
)
2
−
2
d
x
Put
x
+
1
x
=
y
⇒
d
y
=
(
1
−
1
x
2
)
d
x
Also, when
x
=
0
⇒
y
=
∞
And when
x
=
∞
⇒
y
=
∞
∴
I
2
=
∫
∞
∞
−
1
y
2
−
2
d
y
=
0
∴
I
=
∫
∞
0
x
2
d
x
1
+
x
4
.....(3)
Adding equations (1) and (3), we get
2
I
=
∫
∞
0
1
+
x
2
1
+
x
4
d
x
=
∫
∞
0
1
x
2
+
1
1
x
2
+
x
2
d
x
=
∫
∞
0
1
x
2
+
1
1
x
2
+
x
2
+
2
−
2
d
x
=
∫
∞
0
1
x
2
+
1
(
x
−
1
x
)
2
+
2
d
x
Put
x
−
1
x
=
t
⇒
d
t
=
(
1
+
1
x
2
)
d
x
2
I
=
∫
∞
−
∞
d
t
t
2
+
2
=
[
1
√
2
t
a
n
−
1
t
√
2
]
∞
−
∞
=
π
√
2
∴
I
=
π
2
√
2
Evaluate
∫
π
4
0
sin
x
+
cos
x
9
+
16
{
1
−
(
sin
x
−
cos
x
)
2
}
d
x
Report Question
0%
1
20
(
log
√
3
)
0%
1
10
(
log
3
)
0%
1
20
(
log
3
)
0%
1
20
(
log
3
−
log
2
)
Explanation
Let
I
=
∫
π
4
0
sin
x
+
cos
x
9
+
16
{
1
−
(
sin
x
−
cos
x
)
2
}
d
x
∴
I
=
∫
π
4
0
sin
x
+
cos
x
25
−
16
(
sin
x
−
cos
x
)
2
d
x
,
Substitute
4
(
sin
x
−
cos
x
)
=
t
⟹
4
(
cos
x
+
sin
x
)
d
x
=
d
t
1
4
∫
0
−
4
d
t
25
−
t
2
I
=
1
4
.
1
2
(
5
)
(
log
|
5
−
t
5
+
t
|
)
0
−
4
=
1
40
[
log
|
5
−
0
5
+
0
|
−
log
|
5
+
4
5
−
4
|
]
=
1
40
[
log
1
−
log
(
1
9
)
]
, (where
log
1
=
0
)
=
1
40
[
log
9
]
=
2
40
log
3
=
1
20
(
log
3
)
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
.
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
Let
d
d
x
(
F
(
x
)
)
=
e
sin
x
x
,
x
>
0
. If
∫
4
1
2
e
sin
x
2
x
d
x
=
F
(
k
)
−
F
(
1
)
, then the possible value of
k
is
Report Question
0%
10
0%
14
0%
16
0%
18
Explanation
Substitute
x
2
=
z
Then
∫
4
1
2
e
sin
x
2
x
d
x
=
∫
16
1
e
sin
z
z
d
z
=
∫
16
1
d
d
z
{
F
(
z
)
}
d
z
=
(
F
(
z
)
)
16
1
=
F
(
16
)
−
F
(
1
)
∴
F
(
16
)
−
F
(
k
)
=
F
(
16
)
−
F
(
1
)
Gives
k
=
16
The evaluation of
∫
p
X
p
+
2
q
−
1
−
q
X
q
−
1
X
2
p
+
2
q
+
2
X
p
+
q
+
1
d
x
is
Report Question
0%
−
X
p
X
p
+
q
+
1
+
C
0%
X
q
X
p
+
q
+
1
+
C
0%
−
X
q
X
p
+
q
+
1
+
C
0%
X
p
X
p
+
q
+
1
+
C
Explanation
Let us take
x
m
x
(
p
+
q
)
+
1
=
t
where m is either p or q as seen from the options,
d
t
d
x
=
(
x
(
p
+
q
)
+
1
)
m
x
m
−
1
−
x
m
(
p
+
q
)
x
p
+
q
−
1
x
2
p
+
2
q
+
1
+
2
x
p
+
q
=
(
m
x
(
p
+
q
+
m
−
1
)
+
m
x
m
−
1
−
p
x
p
+
q
+
m
−
1
−
q
x
p
+
q
+
m
−
1
)
x
2
p
+
2
q
+
1
+
2
x
p
+
q
we want powers of
p
+
2
q
−
1
and
q
−
1
, so
m
=
q
,
=
(
q
x
(
p
+
2
q
−
1
)
+
q
x
q
−
1
−
p
x
p
+
2
q
−
1
−
q
x
p
+
2
q
−
1
)
x
2
p
+
2
q
+
1
+
2
x
p
+
q
=
q
x
q
−
1
−
p
x
p
+
2
q
−
1
x
2
p
+
2
q
+
1
+
2
x
p
+
q
=
d
t
d
x
so the integral will be
−
x
q
x
(
p
+
q
)
+
1
Let
d
f
(
x
)
d
x
=
e
sin
x
x
,
x
>
0
. If
∫
4
1
3
e
sin
x
3
x
d
x
=
f
(
k
)
−
f
(
1
)
then one of the possible values of
k
is
Report Question
0%
16
0%
63
0%
64
0%
15
Explanation
Substitute
x
3
=
z
Then
∫
4
1
3
e
sin
x
2
x
d
x
=
∫
16
1
e
sin
z
z
d
z
=
∫
64
1
d
d
z
{
F
(
z
)
}
d
z
=
(
F
(
z
)
)
64
1
=
F
(
64
)
−
F
(
1
)
∴
F
(
64
)
−
F
(
k
)
=
F
(
64
)
−
F
(
1
)
Gives
k
=
64
∫
e
37
1
π
sin
(
π
log
e
x
)
x
d
x
is equal to
Report Question
0%
2
0%
−
2
0%
2
/
π
0%
2
π
Explanation
Let
I
=
∫
e
37
1
π
sin
(
π
log
e
x
)
x
d
x
Substitute
π
log
x
=
t
⇒
π
x
d
x
=
d
t
∴
I
=
∫
37
π
0
sin
t
d
t
=
−
[
cos
t
]
37
π
0
=
−
[
cos
37
π
−
cos
0
]
=
2
∫
2
0
√
2
+
x
2
−
x
d
x
is equal to
Report Question
0%
π
+
1
0%
1
+
π
/
2
0%
π
+
3
/
2
0%
none of these
Explanation
Let
I
=
√
2
+
x
2
−
x
d
t
Put
t
=
2
+
x
2
−
x
⇒
d
t
=
(
1
2
−
x
−
−
x
−
2
(
2
−
x
)
2
)
d
x
∴
I
=
4
∫
√
t
(
t
+
1
)
2
d
t
Put
u
=
√
t
⇒
d
u
=
1
2
√
t
d
t
∴
I
=
8
∫
u
2
(
u
2
+
1
)
2
d
u
=
8
∫
(
2
u
2
+
1
−
1
(
u
2
+
1
)
2
)
d
u
=
4
(
(
u
2
+
1
)
tan
−
1
u
−
u
)
u
2
+
1
+
C
=
4
(
(
t
+
1
)
tan
−
1
(
√
t
)
−
√
t
)
t
+
1
+
C
=
√
2
+
x
2
−
x
(
x
−
2
)
+
4
tan
−
1
(
√
2
+
x
2
−
x
)
+
C
Therefore
∫
2
0
√
2
+
x
2
−
x
d
x
=
[
√
2
+
x
2
−
x
(
x
−
2
)
+
4
tan
−
1
(
√
2
+
x
2
−
x
)
]
2
0
=
[
−
2
+
4
π
2
−
]
−
0
=
2
π
−
2
∫
∞
0
f
(
x
+
1
x
)
.
ln
x
x
d
x
is equal to
Report Question
0%
0
0%
1
0%
1
2
0%
cannot be evaluated
Explanation
I
=
∫
∞
0
f
(
x
+
1
x
)
.
ln
x
x
d
x
Let
x
=
e
t
⇒
d
x
=
e
t
d
t
I
=
∫
∞
−
∞
f
(
e
t
+
e
−
t
)
t
d
t
=
0
since,
f
(
e
t
+
e
−
t
)
t
is an odd function.
Ans: 0
∫
∞
1
log
(
t
−
1
)
t
2
log
t
+
log
(
t
t
−
1
)
d
t
equals
Report Question
0%
1
2
0%
1
3
0%
2
3
0%
None of these
Explanation
∫
∞
1
log
(
t
−
1
)
t
2
log
t
log
t
(
t
−
1
)
d
t
=
∫
1
0
log
(
1
x
−
1
)
log
x
log
(
1
−
x
)
d
x
(By substituting
t
=
1
x
d
t
=
1
x
2
d
x
and
t
=
1
⇔
x
=
1
and
t
=
∞
⇔
x
=
0
)
∴
I
=
∫
1
0
log
(
1
−
x
)
−
log
x
log
x
log
(
1
−
x
)
d
x
=
∫
1
0
(
1
log
x
−
1
log
(
1
−
x
)
)
d
x
=
∫
1
0
(
1
log
(
1
−
x
)
−
1
log
x
)
d
x
=
−
I
∴
I
=
0
[
∵
∫
a
0
f
(
x
)
d
x
=
∫
a
0
f
(
a
−
x
)
d
x
]
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
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.
Let
F
(
x
)
=
f
(
x
)
+
f
(
1
x
)
where
f
(
x
)
=
∫
x
1
log
t
1
+
t
d
t
Then
F
(
e
)
is equal to?
Report Question
0%
1
0%
2
0%
1
/
2
0%
0
Explanation
F
(
x
)
=
∫
x
1
ln
t
1
+
t
d
t
+
∫
1
/
x
1
ln
t
1
+
t
d
t
F
(
x
)
=
∫
x
1
(
ln
t
1
+
t
+
ln
t
(
1
+
t
)
t
)
d
t
=
∫
t
1
ln
t
t
d
t
=
1
2
(
ln
x
)
2
F
(
e
)
=
1
/
2
The value of
∫
π
/
2
0
d
θ
5
+
3
cos
θ
is?
Report Question
0%
tan
−
1
1
2
0%
tan
−
1
1
3
0%
1
2
tan
−
1
1
2
0%
1
3
tan
−
1
1
3
Explanation
Let
I
=
∫
π
/
2
0
d
θ
5
+
3
cos
θ
=
∫
π
/
2
0
d
θ
5
+
3.
1
−
tan
2
θ
2
1
+
tan
2
θ
2
=
∫
π
/
2
0
sec
2
θ
2
d
θ
8
+
2
tan
2
θ
2
=
∫
π
/
2
0
1
2
sec
2
(
θ
2
)
d
θ
4
+
tan
2
(
θ
2
)
=
∫
1
0
d
t
4
+
t
2
[
Put
tan
θ
2
=
t
So,
1
2
sec
2
θ
2
d
θ
=
d
t
]
∴
I
=
=
1
2
[
tan
−
1
t
2
]
1
0
=
1
2
(
tan
−
1
1
2
−
tan
−
1
0
)
=
1
2
tan
−
1
1
2
∫
∞
0
1
1
+
x
n
d
x
,
∀
n
>
1
is equal to?
Report Question
0%
2
∫
∞
0
1
1
+
x
d
x
0%
∫
∞
−
∞
1
1
+
x
n
d
x
0%
∫
∞
1
d
x
(
x
n
−
1
)
1
/
n
0%
∫
1
0
1
(
1
−
x
n
)
1
/
n
Explanation
Let
I
=
∫
∞
0
1
1
+
x
n
d
x
=
∫
0
∞
−
d
x
1
+
x
n
Substitute
x
=
1
y
⇒
d
x
=
−
1
y
2
I
=
∫
∞
0
1
y
2
d
y
1
+
(
1
y
)
n
=
∫
∞
0
y
n
−
1
y
(
1
+
y
n
)
d
y
Substitute
t
n
−
1
+
y
n
⇒
n
t
n
−
1
d
t
=
n
y
n
−
1
d
y
I
=
∫
∞
1
d
t
t
(
t
n
−
t
)
1
/
n
=
∫
1
∞
−
1
t
2
.
d
t
(
1
−
1
t
n
)
1
/
n
Substitute
z
=
1
t
⇒
d
x
=
−
1
t
2
d
t
I
=
∫
1
0
d
z
(
1
−
z
n
)
1
/
n
=
∫
1
0
d
x
(
1
−
x
n
)
1
/
n
If
I
1
=
∫
π
/
2
0
x
sin
x
d
x
and
I
2
=
∫
π
/
2
0
tan
−
1
x
x
d
x
,
then
I
1
I
2
=
Report Question
0%
1
2
0%
1
0%
2
0%
π
2
Explanation
Substitute
x
=
tan
θ
∴
I
2
=
∫
π
/
4
0
θ
tan
θ
.
sec
2
θ
d
θ
or
I
2
=
∫
π
/
4
0
2
θ
2
sin
θ
cos
θ
d
θ
=
∫
π
/
4
0
2
θ
sin
2
θ
d
θ
Now substitute
2
θ
=
t
∴
I
2
=
1
2
∫
π
/
2
0
t
sin
t
d
t
=
1
2
I
1
∴
I
1
I
2
=
2
⇒
(
C
)
I
f
∫
1
−
1
g
(
x
)
1
+
t
2
d
t
=
f
(
x
)
,
w
h
e
r
e
,
g
(
x
)
=
sin
x
, then
f
′
(
π
3
)
equals
Report Question
0%
π
4
0%
does not exist
0%
π
√
3
4
0%
None of these
Explanation
f
(
x
)
=
∫
1
−
1
sin
x
1
+
t
2
d
t
=
sin
x
∫
1
−
1
1
1
+
t
2
d
t
=
sin
x
[
tan
−
1
t
]
1
−
1
=
sin
x
[
π
4
+
π
4
]
=
sin
x
[
π
2
]
∴
f
′
(
x
)
=
π
2
cos
x
⇒
f
′
(
π
3
)
=
π
2
1
2
=
π
4
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.
∫
a
0
x
4
(
a
2
−
x
2
)
1
/
2
d
x
equals
Report Question
0%
π
a
5
32
0%
π
a
6
32
0%
π
a
2
32
0%
None of these
Explanation
I
=
∫
a
0
x
4
√
a
2
−
x
2
d
x
=
∫
a
0
x
5
√
(
a
x
)
2
−
1
d
x
Put
(
a
x
)
2
=
t
⇒
(
−
2
a
2
x
3
)
d
x
=
d
t
I
=
−
∫
1
∞
a
8
t
4
√
t
2
−
1
d
t
=
∫
∞
1
a
8
√
t
2
−
1
√
t
2
d
t
=
a
8
∫
∞
1
√
1
−
1
t
2
d
t
=
π
a
6
32
The value of
∫
tan
x
1
/
e
t
1
+
t
2
d
t
+
∫
cot
x
1
/
e
d
t
t
(
1
+
t
2
)
is
Report Question
0%
1
/
2
0%
1
0%
π
/
4
0%
none of these
Explanation
∫
tan
x
1
/
e
t
1
+
t
2
d
t
+
∫
cot
x
1
/
e
d
t
t
(
1
+
t
2
)
=
∫
tan
x
1
/
e
t
1
+
t
2
d
t
+
∫
cot
x
1
/
e
(
1
t
−
1
1
+
t
2
)
d
t
=
1
2
[
log
(
1
+
t
2
)
]
tan
x
1
/
e
+
[
log
t
]
cot
x
1
/
e
−
[
tan
−
1
t
]
cot
x
1
/
e
=
1
∫
π
/
2
0
sin
x
log
(
sin
x
)
d
x
=
Report Question
0%
log
e
e
0%
log
e
2
0%
log
e
(
e
/
2
)
0%
log
e
(
2
/
e
)
Explanation
I
=
∫
π
/
2
0
sin
x
log
(
sin
x
)
d
x
=
1
2
∫
π
/
2
0
sin
x
log
sin
2
x
d
x
=
1
2
∫
π
/
2
0
sin
x
log
(
1
−
cos
2
x
)
d
x
=
1
2
∫
1
0
log
(
1
−
t
2
)
d
t
where
t
=
cos
x
=
1
2
∫
1
0
[
−
t
2
−
(
t
2
)
2
2
−
(
t
2
)
3
3
−
⋯
]
d
t
=
−
[
1
2.3
+
1
4.5
+
1
6.7
+
⋯
]
=
−
[
(
1
2
−
1
3
)
+
(
1
4
−
1
5
)
+
(
1
6
−
1
7
)
+
⋯
]
=
−
1
+
(
1
−
1
2
+
1
3
−
1
4
+
1
5
−
⋯
)
.
=
−
log
e
e
+
log
e
(
1
+
1
)
=
log
e
(
2
/
e
)
Evaluate :
∫
1
√
2
−
1
√
2
x
8
1
−
x
4
×
[
sin
−
1
(
1
−
2
x
2
)
+
cos
−
1
(
2
x
√
1
−
x
2
)
]
d
x
Report Question
0%
π
[
1
2
log
√
2
+
1
√
2
−
1
+
tan
−
1
1
√
2
−
21
10
√
2
]
0%
π
[
1
2
log
√
2
+
1
√
2
−
1
+
tan
−
1
1
√
2
+
21
10
√
2
]
0%
π
[
1
2
log
√
2
−
1
√
2
−
1
+
tan
−
1
1
√
2
−
21
10
√
2
]
0%
π
[
1
2
log
√
2
−
1
√
2
−
1
+
tan
−
1
1
√
2
+
21
10
√
2
]
Explanation
In
N
r
put
x
=
sin
θ
,
then
sin
−
1
(
1
−
2
sin
2
θ
)
+
cos
−
1
(
2
sin
θ
cos
θ
)
=
sin
−
1
(
cos
2
θ
)
+
cos
−
1
(
sin
2
θ
)
=
π
2
−
cos
−
1
(
cos
2
θ
)
+
π
2
−
sin
−
1
(
sin
2
θ
)
=
π
−
2
θ
−
2
θ
=
π
−
4
sin
−
1
x
∴
I
=
∫
1
√
2
−
1
√
2
x
8
1
−
x
4
[
π
−
4
sin
−
∣
x
]
d
x
=
2
∫
1
√
2
0
π
x
8
1
−
x
4
d
x
+
0
by Prop.V
=
2
π
∫
1
√
2
0
x
8
−
1
+
1
1
−
x
4
d
x
=
2
π
∫
1
√
2
0
[
−
(
x
4
+
1
)
+
1
(
1
−
x
2
)
(
1
+
x
2
)
]
d
x
=
2
π
∫
1
√
2
0
−
(
x
4
+
1
)
+
1
2
{
1
1
−
x
2
+
1
1
+
x
2
}
d
x
=
2
x
[
−
(
x
5
5
+
x
)
+
1
4
log
1
+
x
1
−
x
+
1
2
tan
−
1
x
]
1
/
√
2
0
=
π
[
1
2
log
√
2
+
1
√
2
−
1
+
tan
−
1
1
√
2
−
21
10
√
2
]
Ans: A
If
∫
x
log
2
d
x
√
e
x
−
1
=
π
6
, the value of
x
is
Report Question
0%
4
0%
log
8
0%
log
4
0%
none of these
Explanation
Let
I
=
∫
x
log
2
d
x
√
e
x
−
1
=
π
6
Put
e
x
−
1
=
t
2
⇒
2
tan
−
1
√
e
2
−
1
−
π
2
=
π
6
⇒
√
e
2
−
1
=
tan
π
3
=
√
3
⇒
e
x
=
4
⇒
x
=
log
4
∫
π
/
3
0
cos
θ
3
+
4
sin
θ
d
θ
=
λ
log
3
+
2
√
3
3
then
λ
equals
Report Question
0%
1
2
0%
1
3
0%
1
4
0%
1
8
Explanation
Given
∫
π
/
3
0
cos
θ
3
+
4
sin
θ
d
θ
=
λ
log
3
+
2
√
3
3
....(1)
Consider,
I
=
∫
π
/
3
0
cos
θ
3
+
4
sin
θ
d
θ
Substitute
sin
θ
=
t
⇒
cos
θ
d
θ
=
d
t
I
=
∫
√
3
/
2
0
d
t
3
+
4
t
Substitute
3
+
4
t
=
u
4
d
t
=
d
u
∴
I
=
1
4
∫
3
+
2
√
3
3
d
u
u
=
1
4
[
log
u
]
3
+
2
√
3
3
I
=
1
4
[
log
(
3
+
2
√
3
)
−
log
3
]
⇒
I
=
1
4
log
3
+
2
√
3
3
So, on comparing with (1), we get
λ
=
1
4
The value of
∫
tan
x
1
/
e
t
d
t
1
+
t
2
+
∫
cot
x
1
/
e
d
t
t
(
1
+
t
2
)
is
Report Question
0%
1
2
+
t
a
n
2
x
0%
1
0%
π
/
4
0%
2
π
∫
1
−
1
d
t
1
+
t
2
Explanation
∫
tan
x
1
/
e
t
1
+
t
2
d
t
+
∫
cot
x
1
/
e
d
t
t
(
1
+
t
2
)
=
∫
tan
x
1
/
e
t
1
+
t
2
d
t
+
∫
cot
x
1
/
e
(
1
t
−
1
1
+
t
2
)
d
t
=
1
2
[
log
(
1
+
t
2
)
]
tan
x
1
/
e
+
[
log
t
]
cot
x
1
/
e
−
[
tan
−
1
t
]
cot
x
1
/
e
=
1
The value of the integral
∫
β
α
d
x
√
(
x
−
α
)
(
β
−
x
)
for
β
>
α
, is
Report Question
0%
sin
−
1
α
/
β
0%
π
/
2
0%
sin
−
1
β
/
2
α
0%
π
Explanation
Given :
∫
β
α
d
x
√
(
x
−
α
)
(
β
−
x
)
I
=
∫
β
α
d
x
√
(
x
−
α
)
(
β
−
x
)
=
∫
β
α
d
x
√
−
α
β
+
x
(
β
+
α
)
−
x
2
I
=
∫
β
α
d
x
√
(
α
+
β
)
2
4
−
α
β
−
[
x
−
β
+
α
2
]
2
I
=
∫
β
α
d
x
√
(
β
−
α
)
2
4
−
[
x
−
(
β
+
α
)
2
]
2
we know that,
∫
1
√
a
2
−
x
2
d
x
=
sin
−
1
(
x
a
)
+
c
I
=
sin
−
1
[
x
−
(
β
+
α
2
)
(
β
−
α
2
)
]
β
α
=
sin
−
1
[
(
β
−
α
2
)
(
β
+
α
2
)
]
−
sin
−
1
[
(
α
−
β
2
)
(
β
−
α
2
)
]
I
=
sin
−
1
(
1
)
−
sin
−
1
(
−
1
)
I
=
π
2
+
π
2
=
π
Hence the correct answer is
π
∫
π
/
4
π
/
6
d
x
sin
2
x
is equal to
Report Question
0%
1
2
log
(
−
1
)
0%
log
(
−
1
)
0%
log
3
0%
1
2
log
√
3
Explanation
Let
I
=
∫
1
sin
2
x
d
x
=
∫
csc
2
x
d
x
Put
2
x
=
t
⇒
2
d
x
=
d
t
I
=
1
2
∫
csc
t
d
t
Multiply numerator and denominator by
cot
t
+
csc
t
I
=
1
2
∫
−
−
csc
2
t
−
cot
t
csc
t
cot
t
+
csc
t
d
t
Put
cot
t
+
csc
t
=
u
⇒
(
−
csc
2
t
−
cot
t
csc
t
)
d
t
=
d
u
I
=
−
1
2
∫
1
u
d
u
=
−
1
2
log
u
=
−
1
2
log
(
cot
t
+
csc
t
)
=
−
1
2
log
(
cot
2
x
+
csc
2
x
)
=
−
1
2
log
(
cot
x
)
Hence
∫
π
/
4
π
/
6
1
sin
2
x
d
x
=
−
1
2
[
log
(
cot
x
)
]
π
/
4
π
/
6
=
1
2
log
√
3
Value of
∫
1
0
d
x
(
1
+
x
2
)
√
1
−
x
2
is?
Report Question
0%
π
2
√
2
0%
π
√
2
0%
√
2
π
0%
2
√
2
π
Explanation
Let
I
=
∫
d
x
(
1
+
x
2
)
√
1
−
x
2
Substitute
x
=
1
t
⇒
d
x
=
−
1
t
2
d
t
I
=
∫
−
t
d
t
(
t
2
+
1
)
√
t
2
−
1
Substitute
t
2
−
1
=
u
2
⇒
2
t
d
t
=
2
u
d
u
I
=
−
∫
u
d
u
(
u
2
+
1
)
u
=
−
∫
1
u
2
+
(
√
2
)
2
d
u
=
−
1
√
2
tan
−
1
(
u
√
2
)
=
−
1
√
2
tan
−
1
(
√
2
x
√
1
−
x
2
)
Therefore
∫
1
0
I
d
x
=
π
2
√
2
The value of
∫
1
1
/
2
d
x
x
√
3
x
2
+
2
x
−
1
is?
Report Question
0%
π
/
2
0%
π
/
3
0%
π
/
6
0%
π
/
√
2
Explanation
Let
I
=
∫
1
x
√
3
x
2
+
2
x
−
1
d
x
=
∫
1
x
√
(
√
3
x
+
1
√
3
)
2
−
4
3
d
x
Put
t
=
√
3
x
+
1
√
3
⇒
d
t
=
√
3
d
t
I
=
1
√
3
∫
−
1
(
√
3
−
3
t
)
√
t
2
−
4
3
d
t
=
−
3
∫
1
(
√
3
−
3
t
)
√
t
2
−
4
3
d
t
I
=
3
∫
2
3
√
3
s
2
+
√
3
d
s
=
2
√
3
∫
1
3
s
2
+
1
d
s
=
2
tan
−
1
(
√
3
s
)
=
tan
−
1
(
x
−
1
√
3
x
2
+
2
x
−
1
)
∴
∫
1
1
/
2
1
x
√
3
x
2
+
2
x
−
1
d
x
=
π
6
If
I
=
∫
∞
1
x
2
−
2
x
3
√
x
2
−
1
d
x
, then
I
equals
Report Question
0%
−
1
0%
0
0%
π
/
2
0%
π
−
√
3
Explanation
I
=
∫
∞
1
x
2
−
2
x
3
√
x
2
−
1
d
x
Put Put
x
=
sec
u
⇒
d
x
=
tan
u
sec
u
d
u
I
=
∫
π
2
0
cos
2
u
(
sec
2
u
−
2
)
d
u
=
∫
π
2
0
(
1
−
sin
2
u
)
(
sec
2
u
−
2
)
d
u
=
∫
π
2
0
(
2
sin
2
u
−
tan
2
u
+
sec
2
u
−
2
)
d
u
=
−
∫
π
2
0
d
u
+
2
∫
π
2
0
sin
2
u
d
u
=
−
∫
π
2
0
d
u
−
∫
π
2
0
cos
2
u
d
u
+
∫
π
2
0
d
u
=
−
[
sin
2
u
2
]
π
2
0
=
0
If
0
<
α
<
π
/
2
then the value of
∫
α
0
d
x
1
−
cos
x
cos
α
is
Report Question
0%
π
/
α
0%
π
/
2
sin
α
0%
π
/
2
cos
α
0%
π
/
2
α
Explanation
Let
I
=
∫
α
0
d
x
1
−
cos
x
cos
α
=
∫
α
0
d
x
(
cos
2
x
2
+
sin
2
x
2
)
−
cos
α
(
cos
2
x
2
−
sin
2
x
2
)
=
∫
α
0
d
x
(
1
−
cos
α
)
cos
2
x
2
+
2
cos
2
α
2
sin
2
x
2
=
1
2
∫
α
0
sec
2
α
2
sec
2
x
2
tan
2
α
2
+
tan
2
x
2
d
x
Substitute
tan
x
2
=
t
I
=
∫
tan
α
2
0
sec
2
α
2
tan
2
α
2
+
t
2
d
t
=
sec
2
α
2
cot
α
2
[
tan
−
1
1
tan
α
2
]
tan
α
2
0
=
π
2
sin
α
Value of
∫
∞
a
d
x
x
4
√
a
2
+
x
2
is
Report Question
0%
2
+
√
2
3
a
4
0%
2
−
√
2
3
a
2
0%
2
−
√
2
3
a
4
0%
√
2
+
1
3
a
2
Explanation
Let
I
=
∫
∞
a
1
x
4
√
a
2
+
x
2
d
x
Substitute
x
=
a
tan
t
⇒
d
x
=
a
sec
2
t
d
t
I
=
a
∫
π
2
π
4
cot
3
t
csc
t
a
5
d
t
=
1
a
4
∫
π
2
π
4
cot
t
csc
t
(
csc
2
t
−
1
)
d
t
Substitute
υ
=
csc
t
⇒
d
υ
=
−
cot
t
csc
t
d
t
I
=
1
a
4
∫
1
√
2
(
υ
2
−
1
)
d
υ
=
1
a
2
[
υ
2
3
−
υ
]
1
√
2
=
2
−
√
2
3
a
4
The value of
∫
−
5
−
4
e
(
x
+
5
)
2
d
x
+
3
∫
2
/
3
1
/
3
e
9
(
x
−
2
/
3
)
2
d
x
is
Report Question
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
If
\displaystyle \int _{ 0 }^{ 1 }{ \frac { \sin { t } }{ 1+t } dt } =\alpha
, then the value of the integral
\displaystyle \int _{ 4\pi -2 }^{ 4\pi }{ \frac { \sin { t/2 } }{ 4\pi +2-t } dt }
in terms of
\alpha
is given by
Report Question
0%
2\alpha
0%
-2\alpha
0%
\alpha
0%
-\alpha
Explanation
\displaystyle \int _{ 4\pi -2 }^{ 4\pi }{ \frac { \sin { t/2 } }{ 4\pi +2-t } dt } =\frac { 1 }{ 2 } \int _{ 4\pi -2 }^{ 4\pi }{ \frac { \sin { t/2 } }{ 1+\left( 2\pi -\dfrac { t }{ 2 } \right) } dt }
\displaystyle =2.\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \frac { \sin { \left( 2\pi -u \right) } }{ 1+u } du }
(
Substitute
\displaystyle 2\pi -\frac { t }{ 2 } =u\Rightarrow dt=-2du)
\displaystyle =\int _{ 0 }^{ 1 }{ \frac { \sin { u } }{ 1+u } du } =-\int _{ 0 }^{ 1 }{ \frac { \sin { t } }{ 1+t } dt } =-\alpha
Value of
\displaystyle \int_{0}^{16}\displaystyle \frac{x^{1/4}}{1+x^{1/2}}\: dx
is
Report Question
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)
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
Report Question
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
Report Question
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
Report Question
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
If
\displaystyle I=\int _{8}^{15} \frac{dx}{(x-3)\sqrt{x+1} }
then
I
equals
Report Question
0%
\displaystyle \frac{1}{2}\log \frac {5}{3}
0%
\displaystyle 2 \log \frac{1}{3}
0%
\displaystyle \frac{1}{2}-\log \frac {1}{5}
0%
\displaystyle 2 \log \frac{5}{3}
Explanation
Put
\sqrt{x+1} =t
or
x + 1 =t^{2}
\displaystyle \therefore I=\int _{3}^{4} \frac{2t}{(t^{2}-4) t} dt=\frac{2}{(2)(2)}\log \left. \left | \frac{t-2}{t+2}\right |\right ]_{3}^{4}
\displaystyle =\frac{1}{2} \left[ \log \frac{1}{3}-\log \frac{1}{5}\right]
\displaystyle =\frac{1}{2}\log \frac{5}{3}
37 If
n > 1,
and
\displaystyle I=\int _{0}^{\infty} \frac{dx}{(x+\sqrt{1+x^{2}})^{n}}
then
I
equals
Report Question
0%
\displaystyle \frac{n}{n^{2}-1}
0%
\displaystyle \frac{2n}{n^{2}-1}
0%
\displaystyle \frac{n}{2(n^{2}-1)}
0%
\displaystyle \sqrt{ n^{2}-1}
Explanation
As
x
and
\sqrt{1+x^{2}}
are involved,
put
\sqrt{1-x^{2}} = t - x
or
1 + x^{2}=t^{2} -2tx + x^{2}
\displaystyle \Rightarrow x= \frac{1}{2}\left(t-\frac{1}{t}\right)
\displaystyle \Rightarrow dx = \frac{1}{2} \left (1+\frac{1}{t^{2}}\right )
\displaystyle \therefore I= \int_{0}^{\infty} \frac{1}{2} \left(1+\frac{1}{t^{2}}\right) \frac{dt}{t^{n}}
=\displaystyle \frac{1}{2} \left.\left( \frac{t^{-n+1}}{1-n}-\frac{t^{-n-1}}{n+1}\right) \right ]_{1}^{\infty}
=\displaystyle \frac{1}{2}\left. \left ( \frac{1}{1-n}\frac{1} {t^{n-1}}-\frac{1}{n+1}\frac{1}{t^{n+1}} \right) \right ]_{1}^{\infty}
=\displaystyle 0+\frac{1}{2} \left( \frac{1}{n-1}+\frac{1}{n+1}\right) =\frac{n}{n^{2}-1}
A function
f
is defined by
\displaystyle f(x)=\frac{1}{2^{r-1}},\frac{1}{2r}<x\leq \frac{1}{2^{r-1}},r=1,2,3,.....
then the value of
\displaystyle \int _{0}^{1}f(x)dx
is equal
Report Question
0%
\cfrac 13
0%
\cfrac 14
0%
\cfrac 23
0%
\cfrac 12
Explanation
\displaystyle \int_{0}^{1}f(x) dx =\sum_{r=1}^{\infty} \int_{2^{-r}}^{2^{-(r-1)}} \frac{1}{2^{r-1}}dx
\displaystyle =\sum_{r=1}^{\infty}\frac{1}{2^{r-1}}[2^{-(r-1)}-2^{-r}]
\displaystyle =\sum_{1}^{\infty}2^{-2(r-1)}- \sum _{1}^{\infty} 2^{-2r+1}
\displaystyle =(2^{2}-2)\sum_{1}^{\infty}2^{-2r}=2\cdot \frac{1}{4}\cdot \frac{1}{1-1/4}=\frac{2}{3}\cdot
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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