Q248 / 270 · NDA 2015 (Phase 2) · Mathematics
Consider the integrals
$A=\int_0^\pi \frac{\sin x d x}{\sin x+\cos x} \text { and } B=\int_0^\pi \frac{\sin x d x}{\sin x-\cos x}$
What is the value of $B$ ?
- A$\frac{\pi}{4}$
- Correct answerB$\frac{\pi}{2}$
- C$\frac{3 \pi}{4}$
- D$\pi$
Solution
$\begin{aligned} & B=\int_0^\pi \frac{\sin x}{\sin x-\cos x} \times \frac{(\sin x+\cos x)}{(\sin x+\cos x)} d x \\ & =-\int_0^\pi \frac{\sin ^2 x+\sin x \cos x}{\cos 2 x} d x \\ & =\frac{-1}{2} \int_0^\pi \frac{2 \sin ^2 x}{\cos 2 x} d x-\frac{1}{2} \int_0^\pi \frac{2 \sin x \cos x}{\cos 2 x} d x \\ & =\frac{-1}{2} \int_0^\pi \frac{1-\cos 2 x}{\cos 2 x} d x-\frac{1}{2} \int_0^\pi \tan 2 x d x \\ & =\frac{-1}{2} \int_0^\pi \sec 2 x d x+\frac{1}{2} \int_0^\pi d x-\frac{1}{2} \int_0^\pi \tan 2 x d x \\ & =\frac{-1}{2}\left[\frac{\log |\sec 2 x+\tan 2 x|}{2}\right]_0^\pi+\frac{1}{2}[x]_0^\pi-\frac{1}{2}\left[\frac{\log |\sec 2 x|}{2}\right]_0^\pi \\ & = \\ & \frac{-1}{4}[\log (1+0)-\log (1+0)]+\frac{1}{2}[\pi+0]-\frac{1}{4}[\log (1)-\log (1)]\end{aligned}$
$\begin{aligned} & = \\ & \frac{-1}{4}[\log (1+0)-\log (1+0)]+\frac{1}{2}[\pi+0]-\frac{1}{4}[\log (1)-\log (1)] \\ & =0+\pi / 2-0=\pi / 2\end{aligned}$
Hence, the correct answer is option: 'B'.