Mathematics

# If $\displaystyle \int_{0}^{\infty }\frac{e^{-ax}\sin bx}{x}\:dx= \tan^{-1}\left ( b/a \right )$ then the value of $\displaystyle \int_{0}^{\infty }\frac{\sin ^2ax}{x^2}\:dx$ equals, $a\neq 0$

$\displaystyle \frac{\pi a}{2}$

##### SOLUTION
$\displaystyle \int_{0}^{\infty }\frac{\sin ^2ax}{x^2}\:dx$

let $u=ax$

$dx=\dfrac{du}{a}$

$x=\dfrac{u}{a}$

$\displaystyle a\int_{0}^{\infty }\frac{\sin ^2ax}{u^2}\:dx$

$\displaystyle \Rightarrow a\int^{\infty}_0\left(\dfrac{1}{u^2}-\dfrac{\cos^2u}{u^2}\right)du$.

$\displaystyle \Rightarrow a\left[\dfrac{-1}{u}\right]^{\infty}_0-a\left[\dfrac{-1}{u}.\cos^2 u-\int \left(\dfrac{-1}{u}\right)(-\sin 2u)du\right]^{\infty}_0$

$\Rightarrow +\dfrac{\pi}{2} a-a\sin (2a\cdot 0)$

$\Rightarrow \boxed{\dfrac{\pi \, a }{2}}$

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

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