Mathematics

Let $$\displaystyle f\left ( x \right )=\frac{\sin x}{x},then\:\int_{0}^{\pi /2}f\left ( x \right )f\left ( \frac{\pi }{2}-x \right )dx=$$


ANSWER

$$\displaystyle \frac{2}{\pi }\int_{0}^{\pi }f\left ( x \right )dx$$


SOLUTION
$$\displaystyle l=\int_{0}^{n/2}\frac{\sin x}{x}\frac{\cos x}{\left ( \dfrac{\pi }{2}-x \right )}dx$$

$$\displaystyle \pi =\int_{0}^{\pi/2 }\sin 2x\left [ \frac{1}{x} +\frac{1}{\pi /2-x}\right ]dx$$

$$\displaystyle =\int_{0}^{\pi /2}\frac{\sin 2x}{x}dx+\int_{0}^{\pi /2}\frac{\sin 2x}{\pi /2-x}dx$$

$$\displaystyle =\int_{0}^{\pi /2}\frac{\sin 2x}{x}dx+\int_{0}^{\pi /2}\frac{\sin 2x}{x}dx=4\int_{0}^{\pi /2}\frac{\sin 2x}{2x}dx$$

$$\displaystyle \frac{\pi }{2}=\int_{0}^{\pi }\frac{\sin t}{t}dt$$ [Substitute $$2x=t$$]

$$\displaystyle l=\frac{2}{\pi }\int_{0}^{\pi }\frac{\sin x}{x}dx$$
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Single Correct Medium Published on 17th 09, 2020
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