Test Index

UPSC NDA II 2015 Mathematics Solved Paper

Show Para Hide Para

For the next two(2) items that follow:

Consider the integrals

A=\int\limits_{0}^{\pi} \;\frac{\sin x \, d x}{\sin x+\cos x}

and

B=\int\limits_{0}^{\pi} \;\frac{\sin x \, d x}{\sin x-\cos x}

[NDA II 2015]

© examsnet.com

Question : 98 of 120

Marks: +1, -0

What is the value of B?

$\frac{\pi}{4}$
$\frac{\pi}{2}$
$\frac{3\pi}{4}$
$\pi$

Solution:

B=\int\limits_{0}^{\pi} \frac{\sin x}{\sin x-\cos x} \times \frac{\sin x+\cos x}{\sin x+\cos x} dx

=-\int\limits_{0}^{\pi} \frac{\sin^2 x+\sin x \cos x}{\cos 2x} dx

=-\frac{1}{2}\int\limits_{0}^{\pi} \frac{2\sin^2 x}{\cos 2x} dx - \frac{1}{2}\int\limits_{0}^{\pi} \frac{2\sin x \cos x}{\cos 2x} dx

=-\frac{1}{2}\int\limits_{0}^{\pi} \frac{1-\cos 2x}{\cos 2x} dx - \frac{1}{2}\int\limits_{0}^{\pi} \tan 2x dx

=-\frac{1}{2}\int\limits_{0}^{\pi} \sec 2x dx + \frac{1}{2}\int\limits_{0}^{\pi} dx - \frac{1}{2}\int\limits_{0}^{\pi} \tan 2x dx

=-\frac{1}{2}\left[ \frac{\log|\sec 2x+\tan 2x|}{2} \right]_{0}^{\pi} + \frac{1}{2}\left[ x \right]_{0}^{x} - \frac{1}{2}\left[ \frac{\log|\sec 2x|}{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)]

=0+\frac{\pi}{2}-0=\frac{\pi}{2}

© examsnet.com

Go to Question:

Prev Question

Next Question