Question

If $\int \frac{1}{\sin x+\cos x}dx=k\log |\tan (\frac{x}{2}+\frac{\pi }{8})|+c$, then $k=$

• A. $\sqrt{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $-\sqrt{2}$
• D. $\frac{-1}{\sqrt{2}}$

Answer 1

Answer: (B)

$\frac{1}{\sqrt{2}}$

Given, $\int \frac{1}{\sin x+\cos x}dx$
$\Rightarrow$ $\int \frac{1}{\sqrt{2}}\left(\frac{1}{\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x}\right)dx$
$\Rightarrow$ $\frac{1}{\sqrt{2}}\int \frac{dx}{\sin (x+\frac{\pi }{4})}$
$\Rightarrow$ $\frac{1}{\sqrt{2}}\int \text{cosec}(x+\frac{\pi }{4})dx$
Now , since $\int \text{cosec}xdx=\log \left[\tan \right(\frac{x}{2}\left)\right]+c$
So, $\frac{1}{\sqrt{2}}\int \text{cosec}(x+\frac{\pi }{4})dx=\frac{1}{\sqrt{2}}\log \left[\tan \right(\frac{x}{2}+\frac{\pi }{8}\left)\right]+c$
Hence we have $k=\frac{1}{\sqrt{2}}$ as our answer.