Mathematics

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

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

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

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

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

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

#### Realted Questions

Q1 Single Correct Medium
If $b > a,$ and $\displaystyle I = \int _{a}^{b} \sqrt{\frac{ x-a}{b-x} }dx,$ then $I$ equals
• A. $\pi (b -a)$
• B. $\pi/2$
• C. $2\pi (b-a)$
• D. $\displaystyle \frac{\pi}{2} (b -a)$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\displaystyle \int e^{\tan ^{-1}x}(1+x+x^{2}) d(\cot ^{-1}x)$ is equal to
• A. $\displaystyle -e ^{\displaystyle \tan ^{-1}x}+c$
• B. $\displaystyle e ^{\displaystyle \tan ^{-1}x}+c$
• C. $\displaystyle xe ^{\displaystyle \tan ^{-1}x}+c$
• D. $\displaystyle -xe ^{\displaystyle \tan ^{-1}x}+c$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\int ( 1+3x+3x^2+4x^3+........)dx (|x| <1)$-
• A. $(1+x)^{-1}+c$
• B. $(1+x)^{-2}+c$
• C. None of these
• D. $(1-x)^{-1}+c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle\int^{\pi}_{-\pi}x^3\cos^3xdx=?$
• A. $\pi$
• B. $\dfrac{\pi}{4}$
• C. $2\pi$
• D. $0$

$\int {\dfrac {\cos 2x}{\sin x}}dx$