Mathematics

# $\displaystyle\int^{\dfrac{\pi}{4}}_0\dfrac{x\sin x}{\cos^3x}dx$.

##### SOLUTION
$\displaystyle \int_0^{\pi/4} x \tan x \sec^2x dx$
Integrating byparts, we get
$u = x, u^{'} = 1 , v^{'} = \tan x \sec^2 x , v = \cfrac{\sec^2x}{2}$
$\displaystyle \Bigg[\cfrac{x \sec^2x}{2} - \int \cfrac{\sec^2 x }{2}dx\Bigg]_0^{\pi/4}$
$\displaystyle \Bigg[\cfrac{x \sec^2x}{2} - \cfrac{\tan x}{2}\Bigg]_0^{\pi/4}$
$\implies \cfrac{\pi-2}{4}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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