Question

Solve: $\int \frac{dx}{\sec x+\text{cosec}x}$

• A. $\frac{1}{2}(\cos x-\sin x)-\frac{1}{2\sqrt{2}}\ln \left|\frac{\tan \frac{x}{2}+\sqrt{2}}{\tan \frac{x}{2}+\sqrt{2}}\right|+c$
• B. $\left[\sin x-\cos x-\frac{1}{\sqrt{2}}\log \left|\text{cosec}\left(x+\frac{\pi }{4}\right)-\cot \left(x+\frac{\pi }{4}\right)\right|\right]+c$
• C. $\frac{1}{2}\left[\sin x-\cos x-\frac{1}{2}\log \left|\text{cosec}\left(x+\frac{\pi }{4}\right)-\cot \left(x+\frac{\pi }{4}\right)\right|\right]+c$
• D. $\frac{1}{2}\left[\sin x-\cos x-\log \left|\text{cosec}\left(x+\frac{\pi }{4}\right)-\cot \left(x+\frac{\pi }{4}\right)\right|\right]+c$

Answer 1

The correct option is A

$\frac{1}{2}(\cos x-\sin x)-\frac{1}{2\sqrt{2}}\ln \left|\frac{\tan \frac{x}{2}+\sqrt{2}}{\tan \frac{x}{2}+\sqrt{2}}\right|+c$

Consider, $I=\int \frac{1}{\sec x+cosecx}dx$

$\Rightarrow$ $\int \frac{1}{\frac{1}{\cos x}+\frac{1}{\sin x}}dx=\int \frac{\sin x\cos x}{\sin x+\cos x}dx$

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

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

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

$\Rightarrow$ $I=\frac{1}{2}(\cos x-\sin x)-\frac{1}{2}\int \frac{1}{2\sin \frac{x}{2}\cos \frac{x}{2}+co{s}^2\frac{x}{2}-{\sin }^2\frac{x}{2}}dx$

Let this be $I_1$

$\Rightarrow$ $I_1=\frac{1}{2}\int \frac{1}{2\sin \frac{x}{2}\cos \frac{x}{2}+{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}dx$

Divide Numberator and denominator by ${\cos }^2\frac{x}{2}$

$\Rightarrow$ $I_1=\frac{1}{2}\int \frac{{\sec }^2\frac{x}{2}}{2\tan \frac{x}{2}+1-{\tan }^2\frac{x}{2}}dx$

Put $\tan \frac{x}{2}=t$ $\Rightarrow$ $\frac{1}{2}{\sec }^2\frac{x}{2}dx=dt$

$\Rightarrow$ $I_1=-\int \frac{1}{2t+1-t^2}dt=-\int \frac{1}{t^2-2t-1}dt$

$\Rightarrow$ $I_1=-\int \frac{1}{(t-1{)}^2-{\sqrt{2}}^2}dt=\frac{-1}{2\sqrt{2}}ln\left|\frac{t-1-\sqrt{2}}{t-1+\sqrt{2}}\right|+c$

$\Rightarrow$ $I_1=\frac{-1}{2\sqrt{2}}\ln \left|\frac{\tan \frac{x}{2}-1-\sqrt{2}}{\tan \frac{x}{2}-1+\sqrt{2}}\right|$

$\Rightarrow$ $I=\frac{1}{2}(\cos x-\sin x)-\frac{1}{2\sqrt{2}}\ln \left|\frac{\tan \frac{x}{2}+\sqrt{2}}{\tan \frac{x}{2}+\sqrt{2}}\right|+c$