Question

$I=\int \frac{dx}{\sec x+\text{cosec}x}$, then $I$ is equal to

• A. $\{\left(\sin x+\cos x\right)+\frac{1}{\sqrt{2}}\log \left|\frac{\tan \frac{x}{2}-1-\sqrt{2}}{\tan \frac{x}{2}-1+\sqrt{2}}\right|\}+C$
• B. $\{(\sin x+\cos x)+\frac{1}{\sqrt{2}}\log \left|\frac{\tan \frac{x}{2}-1-\sqrt{2}}{\tan \frac{x}{2}-1+\sqrt{2}}\right|\}+C$
• C. $\left\{\frac{1}{2}[\sin x-\cos x]-\frac{1}{2\sqrt{2}}\log |cosec(x+\frac{\pi }{4})-\cot (x+\frac{\pi }{4}\left)\right|\right\}+C$
• D. None of these

Answer 1

Answer: (C)

$\left\{\frac{1}{2}[\sin x-\cos x]-\frac{1}{2\sqrt{2}}\log |cosec(x+\frac{\pi }{4})-\cot (x+\frac{\pi }{4}\left)\right|\right\}+C$

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

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

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

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

$=\frac{1}{2}\int [\sin x+\cos x-\frac{1}{\sqrt{2}}cosec(x+\frac{\pi }{4})]dx$

$=\frac{1}{2}\int [\sin x+\cos x]dx-[\int \frac{1}{\sqrt{2}}cosec(x+\frac{\pi }{4})]dx$

$=\frac{1}{2}[\sin x-\cos x]-\frac{1}{2\sqrt{2}}\log |cosec(x+\frac{\pi }{4})-\cot (x+\frac{\pi }{4}\left)\right|+C$