Question

Evaluate $\int \frac{dx}{\sec x+cosecx}$

• A. $\frac{1}{2}[\sin x-\cos x-\frac{1}{\sqrt{2}}\ln \tan \left(\frac{x}{2}+\frac{\pi }{8}\right)]+c$
• B. $\frac{1}{2}[\sin x+\cos x-\frac{1}{\sqrt{2}}\ln \tan \left(\frac{x}{2}+\frac{\pi }{8}\right)]+c$
• C. $\frac{1}{2}[\sin x-\cos x+\frac{1}{\sqrt{2}}\ln \tan \left(\frac{x}{2}+\frac{\pi }{8}\right)]+c$
• D. None of these.

Answer 1

The correct option is A $\frac{1}{2}[\sin x-\cos x-\frac{1}{\sqrt{2}}\ln \tan \left(\frac{x}{2}+\frac{\pi }{8}\right)]+c$

$I=\int \frac{dx}{secx+cosecx}dx$
$I=\int \frac{\sin x\cos xdx}{\sin x+\cos x}dx$ ........ $[\because \sec x=\frac{1}{\cos x},\text{cosec}x=\frac{1}{\sin x}]$

$I=\frac{1}{2}\int \frac{2\sin x\cos x}{\sin x+\cos x}dx$
$I=\frac{1}{2}\int \frac{1+2\sin x\cos x-1}{\sin x+\cos x}dx$
$I=\frac{1}{2}\int \frac{{\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x-1}{\sin x+\cos x}dx$

$I=\frac{1}{2}\int \frac{(\sin x+\cos x{)}^2-1}{\sin x+\cos x}dx$
$I=\frac{1}{2}[\int (\sin x+\cos x)dx-\int \frac{1}{\sin x+\cos x}dx]$

$I=\frac{1}{2}[\int (\sin x+\cos x)dx-\int \frac{1}{\sqrt{2}(\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x)}dx]$
$I=\frac{1}{2}[-\cos x+\sin x]-\frac{1}{2\sqrt{2}}\int \frac{1}{\sin (x+\frac{\pi }{4})}dx$
$I=\frac{1}{2}[-\cos x+\sin x]-\frac{1}{2\sqrt{2}}\int \text{cosec}(x+\frac{\pi }{4})dx$
$I=\frac{1}{2}[\sin x-\cos x]-\frac{1}{2\sqrt{2}}\ln \left[\tan (\frac{x}{2}+\frac{\pi }{8})\right]+c$