Indefinite Integration
Integration by Parts
Grade None
Question:
<p>[JEE Main 2019] \(\displaystyle\int e^x\frac{1+\sin x}{1+\cos x}\,dx\) equals</p>
<li>\(e^x\tan\dfrac{x}{2}+C\)</li>
<li>\(e^x\cot\dfrac{x}{2}+C\)</li>
<li>\(e^x\sin\dfrac{x}{2}+C\)</li>
<li>\(e^x\cos\dfrac{x}{2}+C\)</li>
Step-by-Step Solution
Key Concept: Use half-angle: 1+cosx=2cos^2(x/2), sinx=2sin(x/2)cos(x/2). Integrand = eˣ[tan(x/2)/2 + sec^2(x/2)/2]. Use \inteˣ(f+f')=eˣf.
<p>Half-angle substitution: $1+\cos x=2\cos^2\!\tfrac x2,\;\sin x=2\sin\tfrac x2\cos\tfrac x2$.</p>
<p>$$\frac{1+\sin x}{1+\cos x}=\frac{1+2\sin\frac x2\cos\frac x2}{2\cos^2\frac x2}=\frac{1}{2}\sec^2\frac{x}{2}+\tan\frac{x}{2}$$</p>
<p>Let $f(x)=\tan\tfrac x2\Rightarrow f'(x)=\tfrac12\sec^2\tfrac x2$.</p>
<p>So the integrand $=e^x[f(x)+f'(x)]$, and:</p>
<p>$$\int e^x[f(x)+f'(x)]\,dx = e^x f(x)+C = e^x\tan\frac{x}{2}+C$$</p>
<p>Answer: <strong>(A)</strong></p>
Correct Answer: A