Question

$\int \frac{1}{\cos x-\sin x}dx$ equals

• A. $\frac{1}{\sqrt{2}}\log \left|\tan (\frac{x}{2}-\frac{3\pi }{8})\right|+C$
• B. $\frac{1}{\sqrt{2}}\log \left|\cot \left(\frac{x}{2}\right)\right|+C$
• C. $\sqrt{2}\log \left|\tan (\frac{x}{2}-\frac{\pi }{8})\right|+C$
• D. $\frac{1}{\sqrt{2}}\log \left|\tan (\frac{x}{2}+\frac{3\pi }{8})\right|+C$

Answer 1

The correct option is D $\frac{1}{\sqrt{2}}\log \left|\tan (\frac{x}{2}+\frac{3\pi }{8})\right|+C$

$\int \frac{1}{a\cos x-b\sin x}dx$, $a=b=1\left(given\right)$
Let $a=r\cos \theta =1$, $b=r\sin \theta =1$
$\therefore$ $r=\sqrt{2}$

$\theta ={\tan }^{-1}\left(\frac{b}{a}\right)=\frac{1}{\sqrt{2}}\int \frac{1}{\frac{1}{\sqrt{2}}\cos x-\frac{1}{\sqrt{2}}\sin x}dx$

$=\frac{1}{\sqrt{2}}\int \frac{1}{\cos (x+\pi /4)}dx=\frac{1}{\sqrt{2}}\int \frac{1}{\sin (\frac{\pi }{2}+x+\frac{\pi }{4})}dx$

$=\frac{1}{\sqrt{2}}\int \frac{1}{2\sin (\frac{x}{2}+\frac{3\pi }{8})\cos (\frac{x}{2}+\frac{3\pi }{8})}dx$

$=\frac{1}{2\sqrt{2}}\int \frac{{\sec }^2(3\frac{\pi }{8}+\frac{x}{2})}{\tan (\frac{x}{2}+\frac{3\pi }{8})}dx$

$=\frac{1}{2\sqrt{2}}\times 2\log \left|\tan (\frac{x}{2}+\frac{3\pi }{8})\right|+c$

$=\frac{1}{\sqrt{2}}\log \left|\tan (\frac{x}{2}+\frac{3\pi }{8})\right|+c$