#### 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 203525
Subjects 9
Chapters 126
Enrolled Students 111

#### Realted Questions

Q1 Subjective Medium
Integrate the function
$x\sqrt {x+2}$

1 Verified Answer | Published on 17th 09, 2020

Q2 One Word Medium
$\displaystyle \int_{0}^{\infty }\left ( \cot ^{-1}x \right )^{2}dx= \frac{\pi}{k} \log 2$. Find the value of $k$.

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
If $\displaystyle \int \frac{\log (t + \sqrt{1 + t^2})}{\sqrt{1 + t^2}} dt = \frac{1}{2} (g (t))^2 + C$ where C is a constant, then $g(2)$ is equal to
• A. $\displaystyle \frac{1}{\sqrt 5} \log (2 + \sqrt 5)$
• B. $2 \log (2 + \sqrt 5)$
• C. $\displaystyle \frac{1}{2}\log (2 + \sqrt 5)$
• D. $\log (2 + \sqrt 5)$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
If $\displaystyle \int_{0}^{\dfrac{\pi}{3}}\dfrac{\tan \theta}{\sqrt{2k \sec \theta}}d\theta=1-\dfrac{1}{\sqrt{2}},(k>0),$ then the value of k is :
• A. $\dfrac{1}{2}$
• B. $4$
• C. $1$
• D. $2$

Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$