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
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