JEE Mains · Maths · STD 12 - 7.1 indefinite integral
Let \(I(x)=\int \frac{x^2\left(x \sec ^2 x+\tan x\right)}{(x \tan x+1)^2} d x\). If \(I(0)=0\) the \(I\) \(\left(\frac{\pi}{4}\right)\) is equal to
- A \(\log _{ e } \frac{( x +4)^2}{16}-\frac{\pi^2}{4(\pi+4)}\)
- B \(\log _{ e } \frac{(x+4)^2}{16}+\frac{\pi^2}{4(\pi+4)}\)
- C \(\log _e \frac{(x+4)^2}{32}-\frac{\pi^2}{4(\pi+4)}\)
- D \(\log _e \frac{(x+4)^2}{32}+\frac{\pi^2}{4(\pi+4)}\)
Answer & Solution
Correct Answer
(C) \(\log _e \frac{(x+4)^2}{32}-\frac{\pi^2}{4(\pi+4)}\)
Step-by-step Solution
Detailed explanation
\(I(x)=\int \frac{x^2\left(x \sec ^2 x+\tan x\right)}{(x \tan x+1)^2} d x\) Let \(x \tan x +1= t\) \(I=x^2\left(\frac{-1}{x \tan x+1}\right)+\int \frac{2 x}{x \tan x+1} d x\) \(I=x^2\left(\frac{-1}{x \tan x+1}\right)+2 \int \frac{x \cos x}{x \sin x+\cos x} d x\)…
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