Indefinite Integration
Integration by Parts
Grade 12
Question:
<p><strong>Column-I</strong> gives integrals; <strong>Column-II</strong> gives conditions or values. Match correctly.</p>
<ul>
<li>(A) Let \(f(x)=(1+\cos x)^{-1}(1-\sin x+\sin^2 x)\,dx\), find \(f\!\left(\frac\pi2\right)\) if \(f\) is cont. in domain → rational</li>
<li>(B) Let \(g(x)=\displaystyle\int\frac{1}{1+e^{2x}}\,dx\) and \(g(0)=0\), then \(g\!\left(\frac12\right)\) → irrational</li>
<li>(C) Let \(k(x)=\displaystyle\int\frac{1}{(x^2+1)}\,dx\) and \(k(0)=0\) → value is \(k\!\left(\dfrac{\pi}{4}\right)\)</li>
<li>(D) If \(\displaystyle\int f(x)\,dx = \ln(x+e^x)\) and \(f(0)=0\) → \(f\) is prime</li>
</ul>
<li>A→P, B→Q, C→R, D→S</li>
<li>A→P, B→S, C→Q, D→R</li>
<li>A→Q, B→P, C→S, D→R</li>
<li>A→R, B→P, C→Q, D→S</li>
Step-by-Step Solution
Key Concept: Evaluate each integral individually and match to the given Column-II properties (rational/irrational/value).
<p>(A) \(\int\frac{1-\sin x+\sin^2 x}{1+\cos x}\,dx\): Use Weierstrass \(t=\tan(x/2)\) or simplify using half-angle identities. Result is rational at \(x=\pi/2\) → P.</p>
<p>(B) \(\int\frac{dx}{1+e^{2x}}\): Let \(u=e^x\Rightarrow du=e^x dx\Rightarrow dx=du/u\). Result involves \(\ln\) giving an irrational value → S.</p>
<p>(C) \(\int\frac{dx}{x^2+1}=\tan^{-1}x\), at \(x=\tan(\pi/4)=1\): \(\tan^{-1}(1)=\pi/4\) → Q.</p>
<p>(D) If \(\int f(x)\,dx=\ln(x+e^x)\Rightarrow f(x)=\frac{1+e^x}{x+e^x}\), check \(f(0)=2/1=2\neq 0\)... actually \(f(0)=\frac{1+1}{0+1}=2\) so \(f\) is not zero at origin → R.</p>
<p>Combination: A→P, B→S, C→Q, D→R = option <strong>B</strong></p>
Correct Answer: B