Mathematics

# Evaluate the following 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)

As we know

$\displaystyle \int_{a}^b f(x) \ dx= \int_{a}^b f(a+b-x) \ dx$

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

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

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

$\Rightarrow 2I=\dfrac{\pi}{4}\displaystyle\int_{0}^{1}\dfrac{1}{(1-t^{2})+t^{2}}dt$,                   where $t=\sin^{2}x\Rightarrow \dfrac{dt}{2}=sinx\,cosx\,dx$

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

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

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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#### Realted Questions

Q1 Single Correct Medium
$\displaystyle\frac{(x+1)^{2}}{x^{3}+x}=\frac{A}{x}+\frac{Bx+C}{x^{2}+1}\Rightarrow \sin^{-1}[\frac{A}{C}]=$
• A. $\displaystyle\frac{\pi }{4}$
• B. $\displaystyle\frac{\pi }{3}$
• C. $\displaystyle\frac{\pi }{2}$
• D. $\displaystyle\frac{\pi }{6}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
Evaluate: $\displaystyle\int \frac{x}{\sqrt{\left ( 1-x^{2} \right )}\cos ^{2}\sqrt{\left ( 1-x^{2} \right )}}dx$
• A. $\displaystyle \tan \sqrt{1-x^{2}}$
• B. $\displaystyle -\tan ({1-x^{2}})$
• C. $\displaystyle -\sec^{2} \sqrt{1-x^{2}}$
• D. $\displaystyle -\tan \sqrt{1-x^{2}}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
Evaluate the integral
$\displaystyle \int_{\alpha}^{\pi/2 - \alpha}$ $\displaystyle \frac{ {c} {o} {t} {x}}{ {t} {a} {n} {x}+ {c} {o} {t} {x}} {d} {x}$
• A. $\displaystyle \frac{\pi}{2}$
• B. $\displaystyle \frac{\pi}{3}$
• C. $\displaystyle \frac{\pi}{8}$
• D. $\displaystyle \frac{\pi}{4}-\alpha$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
Let $\displaystyle\int _{ 0 }^{ 1 }{ \dfrac { { e }^{ t }dt }{ 1+t } }$ then $\displaystyle \int _{ a-1 }^{ a }{ \dfrac { { e }^{ t }dt }{ t-a-1 } }$
• A. $Ae^ {-a}$
• B. $-Ae^ {-a}$
• C. None of these
• D. $Ae^ {a}$

Let $\displaystyle I_{1}=\int_{0}^{1}(1-x^{2})^{1/3} dx$  &  $\displaystyle I_{2}=\int_{0}^{1}(1-x^{3})^{1/2} dx$