Mathematics

# Let $\displaystyle f\left ( x \right )=\frac{\sin 2x \cdot \sin \left ( \dfrac{\pi }{2}\cos x \right )}{2x-\pi }$Then answer the following question.$\displaystyle \int_{0}^{\pi }x^{2}f\left ( x \right )dx$

##### ANSWER

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

##### SOLUTION
Since $\displaystyle \int_{0}^{\pi }f\left ( x \right )dx=\displaystyle \int_{0}^{\pi /2}\left ( f\left ( x \right )+f\left ( \pi -x \right ) \right )dx$

$\therefore$   $\displaystyle \int x^{2}f\left ( x \right )dx=\pi \displaystyle \int_{0}^{\pi /2}\sin 2x\sin \left ( \frac{\pi }{2}\cos x \right )dx$

$\displaystyle =2\pi \displaystyle \int_{0}^{1}t\sin \left ( \frac{\pi }{2}t \right )dx$ (putting $\cos x=t$)

$\displaystyle =2\pi \left [ t\cos \frac{\pi }{2}t\times \frac{2}{\pi } \right ]_{0}^{1}-2\pi \displaystyle \int_{0}^{1}\left ( \cos \frac{\pi }{2}t \right )\left ( -\frac{2}{\pi } \right )dt$

$\displaystyle =0+\frac{2}{\pi }\left ( 2\pi \right )\left ( \sin \frac{\pi }{2}t \right )_{0}^{1}\times \frac{2}{\pi }$

$\displaystyle =\frac{8}{\pi }$

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

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