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BITSAT 2024Paper 2
Exam
BITSAT
Year
2024
Paper
Paper 2
Subject
Mathematics
Chapter
Definite Integration

BITSAT 2024 · Paper 2 · Mathematics · Definite Integration

The integral ∫ (x^2(x ^2 x+ x))/((x x+1)^2) d x is equal to

The integral ∫x2(xsec⁡2x+tan⁡x)(xtan⁡x+1)2dx\int \frac{x^2\left(x \sec ^2 x+\tan x\right)}{(x \tan x+1)^2} d x is equal to

Options

  1. A−x2xtan⁡x+1+c-\frac{x^2}{x \tan x+1}+c
  2. B2log⁡e∣xsin⁡x+cos⁡x∣+c2 \log _e|x \sin x+\cos x|+c
  3. C−x2xtan⁡x+1+2log⁡e∣xsin⁡x+cos⁡x∣+c-\frac{x^2}{x \tan x+1}+2 \log _e|x \sin x+\cos x|+c
  4. Dx2x2tan⁡x−1−2log⁡e∣xsin⁡x+cos⁡x∣+c\frac{x^2}{x^2 \tan x-1}-2 \log _e|x \sin x+\cos x|+c

Correct answer

C. −x2xtan⁡x+1+2log⁡e∣xsin⁡x+cos⁡x∣+c-\frac{x^2}{x \tan x+1}+2 \log _e|x \sin x+\cos x|+c

Step-by-step solution

(c) We note that ddx(xtan⁡x+1)=xsec⁡2x+tan⁡x∴\frac{d}{d x}(x \tan x+1)=x \sec ^2 x+\tan x\therefore integrating by parts with x2x^2 as first function, we get I=∫x2xsec⁡2x+tan⁡x(xtan⁡x+1)2dx=x2(−1xtan⁡x+1)−∫2x(−1xtan⁡x+1)dx=−x2xtan⁡x+1+2∫xcos⁡xxsin⁡x+cos⁡xdx=−x2xtan⁡x+1+2log⁡e∣xsin⁡x+cos⁡x∣+c(∵ddx(xsin⁡x+cos⁡x)=xcos⁡x)\begin{aligned} & I=\int x^2 \frac{x \sec ^2 x+\tan x}{(x \tan x+1)^2} d x \\ & =x^2\left(-\frac{1}{x \tan x+1}\right)-\int 2 x\left(-\frac{1}{x \tan x+1}\right) d x \\ & =-\frac{x^2}{x \tan x+1}+2 \int \frac{x \cos x}{x \sin x+\cos x} d x \\ & =-\frac{x^2}{x \tan x+1}+2 \log _e|x \sin x+\cos x|+c \\ & \quad\left(\because \frac{d}{d x}(x \sin x+\cos x)=x \cos x\right) \end{aligned}
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