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JEE Main 2017 · Mathematics · Indefinite and Definite Integrals · hard

Evaluate ∫ e^x (1+sin x)/(1+cos x) dx, where C denotes an arbitrary constant.

Single correct option

Evaluate ∫ex 1+sin⁡x1+cos⁡x dx\displaystyle\int e^{x}\,\frac{1+\sin x}{1+\cos x}\,dx, where CC denotes an arbitrary constant.
  1. A. excot⁡x2+Ce^x\cot\dfrac x2+C
  2. B. extan⁡x+Ce^x\tan x+C
  3. C. extan⁡x2+Ce^x\tan\dfrac x2+C
  4. D. exsec⁡x2+Ce^x\sec\dfrac x2+C
Show answer and solution

Answer: (C) extan⁡x2+Ce^x\tan\dfrac x2+C

Write 1+sin⁡x1+cos⁡x=11+cos⁡x+sin⁡x1+cos⁡x=12sec⁡2x2+tan⁡x2\dfrac{1+\sin x}{1+\cos x}=\dfrac{1}{1+\cos x}+\dfrac{\sin x}{1+\cos x}=\dfrac12\sec^2\dfrac x2+\tan\dfrac x2, using 1+cos⁡x=2cos⁡2x21+\cos x=2\cos^2\frac x2 and sin⁡x=2sin⁡x2cos⁡x2\sin x=2\sin\frac x2\cos\frac x2. With f(x)=tan⁡x2f(x)=\tan\frac x2 we have f′(x)=12sec⁡2x2f'(x)=\frac12\sec^2\frac x2, so the integrand is ex(f(x)+f′(x))e^x\left(f(x)+f'(x)\right). Since ∫ex(f+f′)dx=exf(x)+C\int e^x\left(f+f'\right)dx=e^xf(x)+C, the integral equals extan⁡x2+Ce^x\tan\dfrac x2+C. Differentiating the answer returns the integrand, which confirms it.
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Original question written in the JEE Main 2017 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake