#### Passage

Let $\displaystyle f\left ( x \right )=\frac{\sin 2x \cdot \sin \left ( \dfrac{\pi }{2}\cos x \right )}{2x-\pi }$

Mathematics

# $\int_{0}^{\pi }xf\left ( x \right )dx=$

$\displaystyle \frac{8}{\pi ^{2}}$

##### SOLUTION
Given $\displaystyle =\int_{0}^{\pi }xf\left ( x \right )dx=\dfrac{1}{2}\int_{0}^{\pi }\dfrac{\left ( 2x-\pi +\pi \right )\sin 2x\sin \left ( \dfrac{\pi }{2}\cos x \right )}{2x-\pi }dx$
$=\displaystyle \dfrac{1}{2}\int_{0}^{\pi }\sin 2x\sin \left ( \dfrac{\pi }{2}\cos x \right )dx+\dfrac{\pi }{2}\int_{0}^{\pi }f\left ( x \right )dx$
$=\displaystyle \dfrac{1}{2}\int_{0}^{\pi }\sin 2x\sin \left ( \dfrac{\pi }{2}\cos x \right )dx+0$          (As $\int_{0}^{\pi }f\left ( x \right )dx=0$)
$\displaystyle =\dfrac{2}{\pi }\int_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}\dfrac{2t}{\pi }\sin tdt=\dfrac{4}{\pi ^{2}}\int_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}t\sin tdt=\dfrac{8}{\pi ^{2}}\int_{0}^{\dfrac{\pi }{2}}t\sin tdt=\dfrac{8}{\pi ^{2}}$
(Substituting $\displaystyle \dfrac{\pi }{2}\cos x=t$ then for $x=0$, $\displaystyle t=\dfrac{\pi }{2}$ and $x=\pi$, $\displaystyle t=-\dfrac{\pi }{2}$)

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Single Correct Medium Published on 17th 09, 2020
Mathematics

# $\int_{0}^{\pi }f\left ( x \right )dx=$

$0$

##### SOLUTION
$\displaystyle I=\int_{0}^{\pi }f\left ( x \right )dx=\int_{0}^{\pi }\dfrac{\sin 2x\sin \left ( \dfrac{\pi }{2}.\cos x \right )}{2x-\pi }dx$
$\therefore$   $\displaystyle I=\int_{0}^{\pi }\dfrac{\sin 2\left ( \pi -x \right )\sin \left ( \dfrac{\pi }{2}.\cos \left ( \pi -x \right ) \right )}{2\left ( \pi -x \right )-\pi }dx$
$\displaystyle =\int_{0}^{\pi }\dfrac{\left ( -1 \right )\sin 2x.\sin \left ( \dfrac{\pi }{2}\cos x \right )}{\pi -2x}dx=-I$
$\therefore$   $2I=0$     $\therefore$   $I=0$

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Single Correct Hard Published on 17th 09, 2020
Questions 203550
Subjects 9
Chapters 126
Enrolled Students 334

#### Realted Questions

Q1 Single Correct Medium
$\int _{ 0 }^{ \pi }{ x } In\left( \sin { x } \right) dx=$
• A. $\dfrac { \pi }{ 2 } ln\ 2$
• B. $\dfrac { -{ \pi }^{ 2 } }{ 2 } ln\ 2$
• C. $-\dfrac { \pi }{ 2 } ln2$
• D. $\ ln\ 2$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate $\int \tan^{-1}\left\{\sqrt{\left(\dfrac{1-\cos 2 x}{1+ \cos2 x}\right)}\right\}dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
Evaluate $\displaystyle{\int}^{\pi}_0 \dfrac{x \sin x}{1+\cos^2 x}dx$.

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle \int_{0}^{\pi/2}\frac{\cos^4x}{\cos^4x+\sin^4x}dx=?$
• A. $\dfrac{\pi}{4}$
• B. $\dfrac{\pi}{2}$
• C. $\dfrac{\pi}{8}$
• D. $\pi$

$\displaystyle \int \frac{\cos x}{5-3\cos x}dx=-\frac{1}{3}x+\frac{k}{6}\tan ^{-1}\left [ 2\tan \frac{x}{2} \right ].$ Find the value of $k$.