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Let\: I(x)=\int \frac{x^2\left(x \sec ^2 x+\tan x\right)}{(x \tan x+1)^2} d x. If\: I(0)=0, then\: I\left(\frac{\pi}{4}\right) is \: equal\: to :

Option: 1

\log _{\mathrm{c}} \frac{(\pi+4)^2}{16}+\frac{\pi^2}{4(\pi+4)}


Option: 2

\log _{\mathrm{c}} \frac{(\pi+4)^2}{32}-\frac{\pi^2}{4(\pi+4)}


Option: 3

\log _{\mathrm{c}} \frac{(\pi+4)^2}{16}-\frac{\pi^2}{4(\pi+4)}


Option: 4

\log _{\mathrm{e}} \frac{(\pi+4)^2}{32}+\frac{\pi^2}{4(\pi+4)}


Answers (1)

best_answer

I(x)=\int \frac{x^2\left(\sec ^2 x+\tan x\right)}{(x \tan x+1)^2} d x

Let \: x \tan x+1=t
I=x^2\left(\frac{-1}{x \tan x+1}\right)+\int \frac{2 x}{x \tan x+1} d x
I=x^2\left(\frac{-1}{x \tan x+1}\right)+2 \int \frac{2 x}{x \tan x+1} d x
I=x^2\left(\frac{-1}{x \tan x+1}\right)+2 \ln |x \sin x+\cos x|+C

As \: I (0) =0 \Rightarrow \mathrm{C}=0

I\left(\frac{\pi}{4}\right)=\ln \left(\frac{(\pi+4)^2}{32}\right)-\frac{\pi^2}{4(\pi+4)}

Posted by

vishal kumar

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