Mathematics

# If $f \left ( x \right )= \dfrac{\sin x}{x}\:\forall \:x\:\in \left [ 0,\pi \right ]$, then $\displaystyle \frac{\pi }{2}\int_{0}^{\frac{\pi}{2}}f \left ( x \right )f \left ( \frac{\pi }{2}-x \right )\:dx$ is equal to

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

##### SOLUTION
Let, $I=\displaystyle \frac{\pi}{2}\int_{0}^{\frac{\pi}{2}}f(x)f\left ( \frac{\pi}{2}-x \right )dx$
$=\displaystyle \frac{\pi}{2}\int_{0}^{\frac{\pi}{2}}\frac{\sin x}{x}\frac{\cos x}{\dfrac{\pi}{2}-x}dx=\displaystyle\frac{\pi}{2}\int_{0}^{\frac{\pi}{2}}\frac{\sin2x}{x\left ( \pi -2x \right )}dx$
$=\displaystyle\frac{\pi}{2}\int_{0}^{\pi}\frac{\sin z}{z(\pi-z)}dz$ $($Let $2x=z\Rightarrow 2dx = dz)$
$=\displaystyle\frac{1}{2}\int_{0}^{\pi}\sin z\left ( \frac{1}{z}+\frac{1}{\pi-z} \right )dz$
$=\displaystyle\frac{1}{2}\left(\int_{0}^{\pi}\frac{\sin z}{z}dz+\int_{0}^{\pi}\frac{\sin z}{\pi-z}dz\right)=\displaystyle\int_{0}^{\pi}\frac{\sin z}{z}dz$

$\because\displaystyle\int_{0}^{a}f(x)=\int_{0}^{a}f(a-x)dx$

$\therefore\dfrac{\pi}2\displaystyle\int_0^{\frac{\pi}2}f(x)f\left(\dfrac{\pi}2 -x\right)\>dx =\displaystyle\int_{0}^{\pi}f(x)dx$

Hence, option C.

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

#### Realted Questions

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$\int{(x+5)^2} dx$ . Integrate this using fundamental  properties of indefinite integral.

1 Verified Answer | Published on 17th 09, 2020

Q2 Multiple Correct Hard
If $\displaystyle \int xe^{-5x^2} \: sin \: 4x^2 dx = Ke^{-5x^2} (A \: sin \: 4x^2 + B \: cos \: 4x^2) + C$. Then
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Q3 Single Correct Hard
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