Mathematics

# Evaluate the  integral :$\displaystyle\int_{0}^{\pi/2}\dfrac{x\sin x\cos x}{\sin^{4}x+\cos^{4}x}\ dx$

##### SOLUTION
We have,
$I=\displaystyle\int_{0}^{\pi/2}\dfrac{x\sin x\cos x}{\sin^{4}x+\cos^{4}x}\ dx$............$(i)$

$\Rightarrow I=\displaystyle\int_{0}^{\pi/2}\dfrac{\left(\dfrac{\pi}{2}-x\right)\cos x\sin x}{\cos^{4}x+\sin^{4}x}dx$.........$\left (\because \int_{a}^{b}f(x) \ dx=\int_{a}^{b}f(a+b-x) \right ) dx$............$(ii)$

Adding $(i)$ and $(ii)$ , we get

$2I=\dfrac{\pi}{2}\displaystyle\int_{0}^{\pi/2}\dfrac{\sin x\cos x}{\cos^{4}x+\sin^{4}x}dx$

Let $t=\sin^{2}x\Rightarrow dt=2 \ \sin x \ \cos x \ dx$

also,
$\sin^2 {\dfrac {\pi} {2}}=1=upper limit$

$\sin^2 {0}=0=lower limit$

Replacing we get,
$2I=\dfrac{\pi}{4}\displaystyle\int_{0}^{1}\dfrac{1}{(1-t^{2})+t^{2}}dt$, where

$2I=\dfrac{\pi}{8}\displaystyle\int_{0}^{1}\dfrac{1}{\left(t-\dfrac{1}{2}\right)^{2}+\left(\dfrac{1}{2}\right)^{2}}dt=\dfrac{\pi}{8}\times 2\left[\tan^{-1} (2t-1)\right]_{0}^{1}$

$\Rightarrow I=\dfrac{\pi}{8}\left(\dfrac{\pi}{4}+\dfrac{\pi}{4}\right)=\dfrac{\pi^{2}}{16}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Subjective Medium
$\int_{0}^{\cfrac{\pi }{2}}{\left( \dfrac{1}{1+\cot x} \right)}dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
Solve: $\displaystyle\int \dfrac{\sin 2x}{\sin^4 x+\cos^4 x} dx=$
• A. $\tan^{-1}\left(\dfrac{1}{2}\tan x\right)$
• B. $\tan^{-1}\left(\sqrt{\tan x}\right)$
• C. $\tan^{-1} (2\tan x)$
• D. $\tan^{-1}(\tan^2 x)$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
The value of $\int\limits_0^\pi {\frac{{{e^{\cos x}}}}{{{e^{\cos x}} + {e^{ - \cos x}}}}dx}$
• A. $\pi$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{5}$
• D. $\frac{\pi }{2}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
Solve $\displaystyle \int { { x }^{ 2 }.\sin {2 x }\ dx }$
• A. $\dfrac { { x }^{ 2 }\cos { 2x } }{ 2 } +\dfrac { 2x\sin { 2x } }{ 4 } +\dfrac { 2\cos { 2x } }{ 8 } +c$
• B. $\dfrac { -{ x }^{ 2 }\cos { 2x } }{ 2 } -\dfrac { x\sin { 2x } }{ 2 } +\dfrac { \sin{ 2x } }{ 8 } +c$
• C. $None$
• D. $\dfrac { {- x }^{ 2 }\cos { 2x } }{ 2 } +\dfrac { 2x\sin { 2x } }{ 4 } +\dfrac { 2\cos { 2x } }{ 8 } +c$

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Hard
If $\displaystyle \int_{1/4}^{x}\displaystyle \frac{dt}{\sqrt{t-t^{2}}}= \displaystyle \frac{\pi }{6}$, then $x$ equals
• A. $1/3$
• B. $1$
• C. none of these
• D. $1/2$