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Question Number 122822 by bemath last updated on 19/Nov/20

	Answered by liberty last updated on 20/Nov/20
	$I=∫ x tan x sec^2 x dx = ∫ x tan x d(tan x) by D.I method → { ((u=x→du=dx)),((v=∫tan x d(tan x)=(1/2)tan^2 x)) :} I=(1/2)x.tan^2 x−(1/2)∫tan^2 x dx I=(1/2)x.tan^2 x−(1/2)∫ (sec^2 x−1)dx I=(1/2)x.tan^2 x−(1/2)tan x+(1/2)x+c thus ∫_0 ^(π/4) ((x.sin x)/(cos^3 x)) dx = [ (1/2)x.tan^2 x−(1/2)tan x+(1/2)x+c ]_0 ^(π/4) = ((2π)/8)−(1/2)= ((π−2)/4).▲$
	$I = \int x \tan x \sec^{2} x d x = \int x \tan x d (\tan x)$ $b y D . I m e t h o d \to {\begin{cases} u = x \to d u = d x \\ v = \int \tan x d (\tan x) = \frac{1}{2} \tan^{2} x \end{cases}$ $I = \frac{1}{2} x . \tan^{2} x - \frac{1}{2} \int \tan^{2} x d x$ $I = \frac{1}{2} x . \tan^{2} x - \frac{1}{2} \int (\sec^{2} x - 1) d x$ $I = \frac{1}{2} x . \tan^{2} x - \frac{1}{2} \tan x + \frac{1}{2} x + c$ $t h u s \underset{0}{\int^{π / 4}} \frac{x . \sin x}{\cos^{3} x} d x = {[\frac{1}{2} x . \tan^{2} x - \frac{1}{2} \tan x + \frac{1}{2} x + c]}_{0}^{π / 4}$ $= \frac{2 π}{8} - \frac{1}{2} = \frac{π - 2}{4} . ▴$
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