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CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 3 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Indefinite Integrals
Quiz 3
∫
cos
x
log
(
cos
x
)
d
x
=
sin
x
log
(
cos
x
)
+
log
|
sec
x
+
tan
x
|
+
f
(
x
)
+
c
, then
f
(
x
)
=
Report Question
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−
sin
x
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cos
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tan
x
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cot
x
lf
∫
f
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d
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=
g
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x
)
then
∫
x
3
f
(
x
2
)
d
x
=
Report Question
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1
2
{
x
2
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d
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1
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{
x
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d
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Explanation
∫
x
3
f
(
x
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d
x
1
2
∫
x
2
f
(
x
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d
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x
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=
k
1
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∫
k
f
(
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d
k
1
2
[
k
∫
f
(
k
)
d
k
−
∫
(
∫
f
(
k
)
d
k
)
d
k
]
=
1
2
[
k
g
(
k
)
−
∫
g
(
k
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d
k
]
∫
f
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d
x
=
∫
f
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d
k
=
g
=
1
2
[
x
2
g
(
x
2
)
−
∫
g
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x
2
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d
x
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]
=
1
2
[
x
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g
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x
2
)
−
∫
g
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x
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d
x
2
]
∫
{
f
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x
)
g
″
(
x
)
−
f
″
(
x
)
g
(
x
)
}
d
x
=
Report Question
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f
(
x
)
g
′
(
x
)
−
g
(
x
)
f
′
(
x
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+
c
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g
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c
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g
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0%
f
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g
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c
Explanation
\displaystyle \int [f\ cx)g^{"}(x)-f^{"}(x)g(x)]dx
\displaystyle \int f(x)g^{''}(x)+f^{'}(x)g^{'}(x)dx
\displaystyle -\int [f^{'}(x)\ g^{'}\ (x)+f^{''}(x)g(x)]dx
\displaystyle \int d[f\ cx)g^{'}(x)]-\int d(f^{'}(x)g(x)]
\displaystyle =\ f(x)g^{'}(x)-f^{'}(x)\ g(x)+c
\displaystyle \int [f(x)g^{''}(x)-f^{''}(x)g(x)]dx
\displaystyle =f(x)g^{'}(x)-f^{'}(x)g(x)+c
\displaystyle \int (e^\sqrt[3]{x}dx)=
Report Question
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x^{2/3}-2 x^{1/3} +2+c
0%
(x^{2/3} - 2x^{1/3} +2)\exp(\sqrt[3]{x})+c
0%
3(x^{2/3} - 2x^{1/3}+2)\exp(\sqrt[3]{x})+c
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2(x^{2/3} - 2x^{1/3}+2)\exp(\sqrt[3]{x})+c
Explanation
I= \int { { e }^{ \sqrt [ 3 ]{ x } }dx }
Put
\sqrt [ 3 ]{ x } =t
\displaystyle \dfrac { 1 }{ 3 } { x }^{ -2/3 }dx=dt
\displaystyle dx=\dfrac { 3 }{ { t }^{ -2 } } dt
So,
I=3\int { { { t }^{ 2 }e }^{ t }dt }
=3[{ { t }^{ 2 }e }^{ t }-2\int { t{ e }^{ t }dt } ]+c
=3{ { t }^{ 2 }e }^{ t }-6[t{ e }^{ t }-\int { { e }^{ t }dt } ]+c
=3{ { t }^{ 2 }e }^{ t }-6t{ e }^{ t }+6{ e }^{ t }+c
=3(x^{2/3} - 2x^{1/3}+2)e^{\sqrt[3]{x}}+c
\displaystyle \int\sqrt{x}.\log xdx=
Report Question
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\displaystyle \frac{2}{3}x^{3/2}.\log x-\frac{4}{9}x^{3/2}+c
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\displaystyle \frac{2}{3}x^{3/2}.\log x+x^{3/2}+c
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x^{3/2}.(\displaystyle \log x-\frac{2}{3})+c
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\displaystyle \frac{2}{5}x^{3/2}(\log x+1)+c
Explanation
\int \sqrt{x}\ logx\ dx
=log\ x\ \int \sqrt{x}\ dx
-\int 1/x\ \left ( \int \sqrt{x}\ dx \right )dx
=2/3\ logx\ x\ ^{3/2}-2/3\int \dfrac{x^{3/2}}{x}dx
2/3\ logx(x\ ^{3/2})-2/3\int x\ ^{1/2}\ dx
2/3\ logx(x\ ^{3/2})-4/9\ x\ ^{3/2}+c
2/3\ x^{3/2}\ log\ x\ 4/9\ x^{3/2}+c
\int \sqrt{x}\ logx\ dx=2/3\ x^{3/2}logx-4/9\ x^{3/2}+c
lf
f(x)dx =g(x) +c
, then
\displaystyle \int f^{-1}(x)dx
is equal to
Report Question
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xf^{-1}(x) +c
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f[g^{-1}(x)]+c
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x f^{-1}(x)-g[f^{-1}(x)]+c
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g^{-1}(x) +c
Explanation
\int f(x)dx=g(x)+c
\Rightarrow \frac{dg(x)}{dx}=f(x)
\int f^{-1}(x)dx
let
x=f(y)
\Rightarrow \int ydf(y)
=y\int df(y)-\int 1[\int df(y)]dy
=yf(y)-\int f(y)dy
=yf(y)-g(y)+c
xf^{-1}(x)-g(f^{-1}(x))+c
So,
\int f^{-1}(x)dx=xf^{-1}(x)-g(f^{-1}(x))+c
Evaluate
\displaystyle \int e^{\sin x}\sin 2xdx
Report Question
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2e^{\sin x}(\sin x+1)+c
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e^{\sin x}(\sin x+2)+c
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e^{\sin x}(2\sin x-2)+c
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e^{\sin x}(3\sin x -2)+c
Explanation
\int e ^{\sin x} \cdot 2 \sin x \cos x dx
=2\int e ^{\sin x} \cdot \sin x d(\sin x)
Take
\sin x=t
=2\int e^{t} \cdot t dt
We know,
\displaystyle \int u.v \ dx=u\int v\ dx
\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx
Now by integrating the functions by parts, by taking
t
as
u
and
e^t
as
v
, we get
=2(t e ^{t}-e ^{t})
=2\ e\ ^{t}(t-1)
Put the value of
t
=2e^{\sin x}(\sin x-1)=e^{\sin x}(2\sin x-2)
\Rightarrow \int e\ ^{\sin x}\sin \ 2x\ dx=e^{\sin x}(2\sin x-2)+c
\displaystyle \int x\tan x\sec^{2}xdx=
Report Question
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\displaystyle \frac{1}{2}[x\tan^{2} x-tanx +x]+c
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\frac{1}{2}[x\tan^{2} x-tanx -x]+c
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\displaystyle \frac{1}{2}[x\tan^{2}x+ tanx-x]+c
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x\tan^{2} x-tanx+x+c
Explanation
L=\int (x\ tanx)sec^{2}x\ dx
=x\ tanx\int sec^{2}x\ dx
-\int \left ( [tanx+x\ sec^{-2}x]\int sec^{2}x\ dx \right )dx
=x\ tanx\ tanx-\int [tan^{2}x+(x\ sec^{2}x\ tan\ x)]dx
L=x\ tan^{2}x-\int tan^{2}x\ dx-\int x\ s\ e\ c^{2}x\ tanx\ dx
2L=x\ tan^{2}x-\int (sec^{2}x-1)dx
2L=x\ tan^{2}x-\ tanx+x+c
L=\dfrac{1}{2}\ [x\ tan^{2}x-tanx+x]+c
\int x\ tanx\ sec^{2}x\ dx=\dfrac{1}{2}\ [x\ tan^{2}x-tanx+x]+c
\displaystyle \int\frac{2x+sin2x}{1+cos2x}dx=
Report Question
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xsin2x+c
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-xtanx+c
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xsecx+c
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xtanx+c
Explanation
\int \dfrac{2x+sin2x}{1+cos2x}dx
\int \dfrac{2x+sin2x}{2cos^{2}x}dx
\int xsec^{2}x^{dx}+\int tanxdx
x\int sec^{2}xdx-\int (\int sec^{2}xdx)dx
+\int tanxdx
x\int sec^{2}xdx-\int tanxdx+\int tanxdx
xtanx+c
\int \dfrac{2x+sin2x}{1+cos2x}=xtanx+c
\displaystyle \int e^{x}(3x^{2}+7x+2)dx=
Report Question
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e^{x}[3x^{2}+x+1] +c
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e^{x}[3x^{2}+7x+1] +c
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e^{x}[3x^{2} - x- l] +c
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e^{x}[3x^{2}+x]+c
Explanation
\int e^{x}(3x^{2}+7x+2)dx
3\int e^{x}x^{2}\ dx+7\ \int\ e^{x}x\ dx+2\int e^{x}dx
\int e^{x}\ x^{2}\ dx=x^{2}e^{x}-2\int xe^{x}\ dx+c
\int xe^{x}=xe^{x}-e^{x}+c
\int e^{x}=e^{x}+c
3[x^{2}\ e^{x}]-6\int xe^{x}\ dx+7\int e^{x}-x\ dx+2e^{x}
3\ x^{2}e^{x}+xe^{x}-e^{x}+e^{x}+2e^{x}+c
=e^{n}[3x^{2}+x+1]+c
\int e^{x}(3x^{2}+7x+2)dx=\ e^{x}[3x^{2}+x+1]+c
Evaluate
\displaystyle \int e^{x}(\log x+\frac{1}{x^{2}})dx
Report Question
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e^{x}
logx
+c
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\displaystyle e^{x}(\log x-\frac{1}{x})+c
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\displaystyle e^{x}(\log x+\frac{1}{x})+c
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\displaystyle \frac{e^{x}}{x^{2}}+c
Explanation
\displaystyle \int e^{x}(log x+\dfrac{1}{x^{2}})dx
\displaystyle \int e^{x}log x dx+\int e^{x}\dfrac{1}{x^{2}}dx
\displaystyle =log x\int e^{x}dx-\int \dfrac{1}{x}e^{x}dx+\int \dfrac{e^{x}}{x^{2}}dx
\displaystyle =log x e^{x}-\int \dfrac{1}{x}e^{x}dx+e^{x}\int \dfrac{1}{x^{2}}dx -\int \left(e^{x}\int \dfrac{1}{x^{2}}dx \right) dx
\displaystyle =log x e^{x}-\int \dfrac{1}{x}e^{x}dx+e^{x}[-\dfrac{1}{x}]+\int e^{x}\dfrac{1}{x}dx
\displaystyle =e^{x}[log x-\dfrac{1}{x}]+c
So,
\displaystyle \int e^{x}(log x+\dfrac{1}{x^{2}})dx = e^{x}[log x-\dfrac{1}{x}]+c
\displaystyle \int x^{3} \cos x^{2}dx=
Report Question
0%
x^{2} \sin x^{2} -\cos x^{2}+c
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\displaystyle \frac{1}{2} [ x^{2} \sin x^{2}+ \cos x^{2}] +c
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\displaystyle \frac{1}{3} [ x^{2} \sin x^{2}+ \cos x^{2}] +c
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x^{2} \sin x^{2}+ \cos x^{2}+c
Explanation
\displaystyle \int x^{3} \cos x^{2}dx
Put
x^{2}=t
xdx=\displaystyle \frac{dt}{2}
=\displaystyle \frac { 1 }{ 2 } \int t\cos tdt
\displaystyle =\frac { 1 }{ 2 } [t\sin { t } -\int { \sin { t } dt } ]
\displaystyle =\frac { 1 }{ 2 } t\sin { t } +\frac { 1 }{ 2 } \cos { t } +c
=\displaystyle \frac{1}{2} [ x^{2} \sin x^{2}+ \cos x^{2}] +c
\displaystyle \int\frac{\log(x^{2}+a^{2})}{x^{2}}dx =
Report Question
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\displaystyle \frac{-\log(x^{2}+a^{2})}{x}+\frac{2}{a}tan^{-1}(\frac{x}{a})+c
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\displaystyle \frac{\log(x^{2}+a^{2})}{x}+\frac{2}{a}tan^{-1}(\frac{x}{a})+c
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\displaystyle \frac{\log(x^{2}+a^{2})}{x}-\frac{2}{a}tan^{-1}(\frac{x}{a})+c
0%
\displaystyle-\frac{\log(x^{2}+a^{2})}{x}-\frac{2}{a}tan^{-l}(\frac{x}{a})+c
Explanation
\int \dfrac{log(x^{2}+a^{2})}{x^{2}}dx
log(x^{2}+a^{2})\int 1/x^{2}dx-\int (\dfrac{2x}{x^{2}+a^{2}}-[\int \dfrac{1}{x^{2}}dx])dx
=log(x^{2}+a^{2})(-1/x)+2\int \dfrac{1}{(x^{2}+a^{2})}dx+c
=log (x^{2}+a^{2})(-1/x)+\dfrac{2}{a}tan^{-1}(x/a)+c
=\dfrac{-log(x^{2}+a^{2})}{x}+\dfrac{2}{a}tan^{-1}(x/a)+c
\int \dfrac{log(x^{2}+a^{2})}{x^{2}}dx=\dfrac{-log(x^{2}+a^{2})}{x}+\dfrac{2}{a}tan^{-1}(x/a)+c
\displaystyle \int\frac{x\sin^{-1}x}{\sqrt{1-x^{2}}}dx
is equal to
Report Question
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x-\sqrt{1-x^{2}}sin^{-1}x+c
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x+\sqrt{1-x^{2}}sin^{-1}x+c
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x+sin^{-1}x+c
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x-sin^{-1}x+c
Explanation
Let
I=\displaystyle \int \dfrac { x\sin ^{ -1 } x }{ \sqrt { 1-x^{ 2 } } } dx
Put
\sin ^{ -1 } x=t
\implies x=\sin t
\dfrac {1}{\sqrt { 1-x^{ 2 } }}dx=dt
I=\int t\sin { t } dt
=-t\cos { t } +\int { \cos { t } dt }
....integration by parts
=-t\cos { t } +\sin { t } +C
=-\sqrt { 1-{ x }^{ 2 } }\sin ^{ -1 } x +x+C
=x-\sqrt { 1-{ x }^{ 2 } }\sin ^{ -1 } x +C
Hence, option 'A' is correct.
\displaystyle \int e^{sin^{-1}x}[1+\frac{x}{\sqrt{1-x^{2}}}]
dx
=
Report Question
0%
xe^{sin^{-1}x}+c
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e^{sin^{-1}x}+c
0%
\displaystyle \frac{1}{\sqrt{1-x^{2}}}e^{\sin^{-1}x}+c
0%
x^{2}e^{sin^{1}X}+c
Explanation
\displaystyle I= \int e^{sin^{-1}x}[1+\frac{x}{\sqrt{1-x^{2}}}]dx
Put
\sin^{-1} x=t
\Rightarrow x=\sin t
\Rightarrow dx=\cos t dt
So,
\displaystyle I=\int e^{t}(\sin t +\cos t)dt
\Rightarrow I = (\sin t +\cos t)e^{t}-\int (\cos t - \sin t)e^{t} dt
\Rightarrow I= (\sin t +\cos t)e^{t} - (\cos t-\sin t )e^{t} -\int e^{t}(\sin t +\cos t)dt
\Rightarrow 2I= 2\sin t e^{t}+C
\Rightarrow I =xe^{\sin^{-1} x}+C
\displaystyle \int e^\sqrt{x}dx=
Report Question
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2 e^\sqrt{x}(\sqrt{x}+1)+c
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2 e^ \sqrt{x} \left( \sqrt{x}-1 \right) + c
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e^ x (x - 1) + c
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e ^x (x + 1) + c
Explanation
I=\int e^{ \sqrt { x } }dx
Put
\sqrt { x } =t
\displaystyle \frac { 1 }{ 2 } { x }^{ -1/2 }dx=dt
\Rightarrow dx=2tdt
So,
I=2\int te^{t}dt
=2[t{ e }^{ t }-\int { { e }^{ t }dt } ]
= 2t{ e }^{ t }-2{ e }^{ t }+C
=2 e^{\sqrt{x} } \left( \sqrt{x}-1 \right)+c
\displaystyle \int tan^{-1}\sqrt{\frac{1-x}{1+x}}d{x}=
Report Question
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\displaystyle \frac{1}{2}[x cos^{-1} x-\sqrt{1-x^{2}}]+c
0%
[xcos^{-1}x-\sqrt{1-x^{2}}]+c
0%
\displaystyle \frac{1}{2}[x cos^{-1} x+\sqrt{1-x^{2}}]+c
0%
x cos^{-1} x+\sqrt{1-x^{2}}+c
Explanation
\displaystyle I =\int tan^{-1}\sqrt{\frac{1-x}{1+x}}d{x}=
Put
x=\cos {\theta}
\Rightarrow dx=-\sin{\theta}d{\theta}
\displaystyle I =\int tan^{-1}(\tan {\frac{\theta}{2}}) (-\sin{\theta})d{\theta}
\Rightarrow \displaystyle I=-\int \frac{\theta}{2} \sin{\theta} d{\theta}
\Rightarrow \displaystyle I=-\frac{1}{2}\int \theta\sin {\theta}d{\theta}
\Rightarrow \displaystyle I=-\frac{1}{2}[{\theta}(-\cos {\theta})+\sin{\theta}]+c
\Rightarrow \displaystyle I=\frac{1}{2}[{\theta}(\cos {\theta})-\sin{\theta}]+c
\displaystyle I= \frac{1}{2}[x cos^{-1} x-\sqrt{1-x^{2}}]+c
(A) :
\displaystyle \int(2x
tan x
\sec^{2}x+\tan^{2}x)dx=x\tan^{2}x+c
(B) :
\displaystyle \int[xf^{'}(x)+f(x)]dx=xf(x)+c
Report Question
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Both A and R are true and R is the correct explanation of A
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Both A and R are true but R is not correct explanation of A
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A is true R is false
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A is false but R is true.
Explanation
B is true
A is in the form of B
So, A and B are true.
and B is correct explanation.
\displaystyle \int\frac{x-\sin x}{1-\cos x}dx=
Report Question
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log |1-Cosx | +c
0%
log | x - sin x | +c
0%
x\displaystyle \tan\frac{x}{2}+c
0%
-x\displaystyle \cot\frac{x}{2}+c
Explanation
\int (\dfrac{x-sin x}{1-cos x})dx=\int \dfrac{x-sin x}{2sin^{2}x/2}dx
=\int (\dfrac{x}{2})co sec^{2}x/2^{dx}-\int cot x/2 dx
=x/2\int cosec^{2}x/2 d(x/2-\int (\int cosec^{2}x/2 dx/2))dx
-\int cot x/2 dx
=-x cot x/2+c+\int cot x/2 dx-\int cot x/2 dx
=-x cot x/2+c
\int (\dfrac{x-sin x}{1-cos x})dx=-x cot x/2+c
\displaystyle \int(\log x)^{2}dx=
Report Question
0%
x[(\log x)^{2}-2
\log x
+2]+c
0%
x[(\log x)^{2}+2
\log x
+2]+c
0%
[(\log x)^{2}-2
\log x
+2]+c
0%
[(\log x)^{2}+2
\log x
+2]+c
Explanation
\displaystyle \int (\log x)_{1}^{2}dx \int (\log x)^{2}-1 dx
\displaystyle (\log x)^{2} \int 1dx- \int \dfrac{2(\log x)}{x}x dx
\displaystyle =(\log x)^{2}[x]-2\int \log x dx
\displaystyle \int \log x=x \log x-x
\displaystyle =x(\log x)^{2}-2[x \log x-x]+c
\displaystyle =x(\log x)^{2}-2x(\log x)+2x+c
\displaystyle =x[(\log x)^{2}-2(\log x)+2]+c
\displaystyle \Rightarrow \int (\log x)^{2}dx=x[(\log x)^{2}-2(\log x)+2]+c
\displaystyle \int x^{n}.\log xdx=
Report Question
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\displaystyle \frac{x^{n+1}}{(n+1)^{2}}(\log x-\frac{1}{(n+1)^{2}})+c
0%
(\displaystyle \log x-\frac{1}{(n+1)^{2}})+c
0%
\displaystyle \frac{x^{n+1}}{(n+1)}(\log x-\frac{1}{(n+1)})+c
0%
\displaystyle \frac{x^{n+1}}{(n+1)}(\log x-\frac{1}{(n+1)^{2}})+c
Explanation
\int x^{n}log x dx
=log x\int x^{n} dx-\int [\dfrac{1}{x}\int x^{n}.dx]dx
=log x\dfrac{x^{n+1}}{n+1}-\dfrac{1}{n+1}\int x^{n}.dx
=log x\dfrac{x^{n+1}}{n+1}-\dfrac{1}{n+1}[\dfrac{x^{n+1}}{(n+1)}]+c
=\dfrac{x^{n+1}}{(n+1)}[logx-\dfrac{1}{(n+1)}]+c
\int x^{n} log x dx=\dfrac{x^{n+1}}{n+1}[logx-\dfrac{1}{(n+1)}]+c
\displaystyle \int\frac{\log(1-x)}{x^{2}}dx
equal to
Report Question
0%
[(1+\displaystyle \frac{1}{x})\log(1-x)-\log x]+c
0%
(1-\displaystyle \frac{1}{x})\log(1-x)-\log x+c
0%
(\displaystyle \frac{1}{x}-1)\log(1-x)+\log x+c
0%
(\displaystyle \frac{1}{x}-1)\log(1-x)-\log x+c
Explanation
\dfrac{\int log(1-x)}{x^{2}}dx
log(1-x)\int 1/x^{2}dx-\int [\dfrac{1}{(1-x)}\int \dfrac{1}{x^{2}}dx ]dx
\dfrac{-log(1-x)}{x}+c-\int \dfrac{1}{(1-x)}-\dfrac{1}{x}dx
=\dfrac{-log(1-x)}{x}-\int [\dfrac{1}{x}+\dfrac{1}{1-x}]dx+c
=\dfrac{-log(1-x)}{x}-log x+log(1-x)+c
=log(1-x)[1-\dfrac{1}{x}]-log x+c
\int \dfrac{log(1-x)}{x^{2}}dx- log(1-x)[1-1/x]-log x+c
\int \sin \sqrt {x}dx =
Report Question
0%
2(Sin\sqrt {x} - \sqrt {x}.\cos \sqrt {x}) + c
0%
2(Sin \sqrt {x} + \sqrt {x}.\cos \sqrt {x}) + c
0%
2 (Sin \sqrt{x} -\dfrac {1}{2} \sqrt {x} .\cos \sqrt {x}) + c
0%
2(Sin \sqrt {x} + \dfrac {1}{2} \sqrt {x} . \cos \sqrt {x}) + c
Explanation
\displaystyle \int 32x^{3}(\log x)^{2}dx
is equal to
Report Question
0%
8x^{4}(\log x)^{2}+c
0%
x^{4}\{8(\log x)^{2}-4\log x+1\}+c
0%
x^{4}\{8(\log x)^{2}-4\log x\}+c
0%
x^{3}\{(\log x)^{2}+2\log x\}+c
Explanation
\int 32\ x^{3}(log\ x)^{2}= 32 \int x^{3}(log\ x)^{2}dx
= 32\ x^{3}\int (log\ x)
\left [ (log\ x)^{2} \right ]x^{3}dx - \int \left ( 2 \dfrac{log\ x}{x}\ \dfrac{x^{4}}{4} \right )dx
= (log x)^{2} \dfrac{-x^{4}}{4} - \int \ ^{1}/_{2}log\ x . x^{3}-dx
= (log\ x)^{2}-\dfrac{x^{4}}{4}-\ ^{1}/_{2} \int log\ x-x^{3}-dx
\int log\ x.x^{3} dx= log\ x.\dfrac{x^{4}}{4}-\int \ ^{1}/_{x} .\dfrac{x^{4^{3}}}{4}.dx
= log\ x.\dfrac{x^{4}}{4}- \ ^{1}/_{16}\ x^{4}+c
32 \left [ (log\ x)^{2}-\dfrac{x^{4}}{4}-log\ x.\dfrac{x^{4}}{8}+\dfrac{1}{32}x^{4} \right ]+c
= x^{4}\left [ 8(log\ x)^{2}-4log\ x+1 \right ]+c
\displaystyle \int x^{3}(\log x)^{2}dx=
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\displaystyle \frac{x^{4}}{4}[(\log x)^{2}-\frac{\log x}{2}+\frac{1}{8}]+c
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\displaystyle \frac{x^{4}}{4}[(\log x)^{2}+\frac{\log x}{2}+\frac{1}{8}]+c
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(\log x)^{2}+2 logx +8+c
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\displaystyle \frac{x^{4}}{4}[(\log x)^{2}-\frac{\log x}{2}-\frac{1}{8}]+c
Explanation
\int x^{3}(log\ x)^{2}dx=(log\ x)^{2}-\int x^{3}dx-\int \left ( 2\dfrac{log\ x}{x}.\dfrac{x^{4}}{4} \right )dx
= (log\ x)^{2}-\dfrac{x^{4}}{4}-\int \dfrac{1}{2} (log\ x)(x^{3})dx
\int (log\ x)(x^{3})dx= log\ x.\dfrac{x^{4}}{4}-\dfrac{x^{4}}{16}+c
\dfrac{x^{4}}{4}\left [ (log\ x)^{2}-\dfrac{log\ x}{2}+\ \dfrac{1}{8} \right ]+c
\int \displaystyle \frac{sin\frac{x}{2}+cos^{2}\frac{x}{2}}{1+cos x}dx=
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\sec\frac{x}{2}+x+c
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\displaystyle \sec\frac{x}{2}+\frac{x}{2}+c
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\displaystyle\sec\frac{x}{2}-\frac{x}{2}+C
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\displaystyle\sec\frac{x}{2}-X+C
Explanation
\int \dfrac{sin\frac{x}{2} +cos^{2}\frac{x}{2}}{1+cos\ x}dx
\int \dfrac{sin\frac{x}{2} +cos^{2}\frac {x}{2}}{2cos^{2}\frac{x}{2}}dx
= \frac{x}{2}\left [ \int tan \frac{x}{2} sec \frac {x}{2}\frac{dx}{2}+\int 1. dx \right ]
sec \frac{x}{2} + x + c
Solve
\displaystyle \int e^{\log x}.\cos xdx
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x \sin x -cos x + c
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\displaystyle \frac{x}{2}\sin x+\cos x+c
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x \sin x + cos x + c
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x \sin x + cos2 x + c
Explanation
We know,
e^{log x}=x
\int e ^{log x}cos x dx= \int x cos x dx
We know,
\displaystyle \int u.v \ dx=u\int v\ dx
\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx
Now by integrating the functions by parts, taking
x
as
u
and
cos x
as
v
, we get
\int x cos x dx = x \int cos x dx - \int 1(\int cos x)dx
=x sin x +cos x +c
\int e ^{log x}cos x dx= x \cdot sin x +cos x+c \cdot
\displaystyle \int\cos x.\log(\cos x)dx=
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sin x log (cos x) - log (cos x) + c
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sin x log (cos x) + sec x + c
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sin x log (cos x) - sin x + log |sec x + tan x| + c
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sinx log (cosx) - sec x + c
Explanation
\int cos\ x\ log(cos\ x)dx.
= log(cos\ x)\int cos\ x\ dx-\int \left [ \dfrac{log(cos\ x)}{dx}-\int (cos\ x)dx \right ]
= log(cos\ x)sin\ x-\int \dfrac{(-sin\ x).(t\ sin\ x)}{cos\ x}dx
= log(cos\ x)sin\ x+\int \dfrac{sin^{2}x}{cos\ x}\ dx
= log(cosx)sinx+\int secx\ dx-\int cosx\ dx
= sin\ x\ log(cos\ x)-sin\ x+log\left | sec x+tan\ x \right |+c
\displaystyle \int x^{3} sin x^{2}dx=
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\displaystyle \frac{1}{2}\sin x^{2}-\frac{1}{2}x^{2}\cos x^{2}+c
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\displaystyle \frac{1}{2}\sin x^{2}+\frac{1}{2}x^{2}\cos x^{2}+c
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\sin x^{2}-x^{2}\cos x^{2}+c
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\displaystyle \sin x^{2}-\frac{1}{2}x^{2}\cos x^{2}+c
Explanation
\int x^{3}sin x^{2}dx
=1/2\int x^{2} sin x^{2}dx^{2}, let x^{2}=t
=1/2\int t sint dt
\int t sint dt=-tcost+sint
=1/2\int tsin dt=1/2[sin t-t cost]+c
=1/2[sin x^{2}-x^{2}cos x^{2}]+c
=1/2sin x^{2}-x^{2}/2 cos x^{2}+c
\int x^{3}sin x^{2} dx=1/2sin x^{2}-\dfrac{x^{2}}{2}cos x^{2}+c
Evaluate
\displaystyle \int x.\sec^{2}x \ dx
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x \tan x + \log |\sec x| + c
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x \tan x - \log |\sec x| + c
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\displaystyle \frac{x}{2}\tan x-\log |\sec x|+c
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\dfrac{x}{2}\tan x+\log|\sec x|+c
Explanation
We know,
\displaystyle \int u.v \ dx=u\int v\ dx
\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx
Now by integrating the functions by parts, taking
x
as
u
and
sec^2 x
as
v
.
\displaystyle \int x. \sec ^{2}xdx =x\int \sec ^{2}xdx-\int 1\left(\int \sec ^{2}xdx\right)dx
\displaystyle =x \tan x-\int \tan x \ dx+c
=x \tan x-\log |\sec x|+c
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
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