Question

Solve $\int \frac{1-\tan x}{1+\tan x}dx=\_\_\_\_+c$

• A. $-{\sec }^2(\frac{\pi }{4}-x)+c$
• B. $\log |\cos x+\sin x|+c$
• C. $\log \left|\sec (\frac{\pi }{4}-x)\right|+c$
• D. $\log \left|\cos (\frac{\pi }{4}-x)\right|+c$

Answer 1

The correct option is B $\log |\cos x+\sin x|+c$

$\int \frac{1-\tan x}{1+\tan x}dx$

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

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

Let $t=\cos x+\sin x$ then $dt=(-\sin x+\cos x)dx=(\cos x-\sin x)dx$

Now $=\int \frac{\cos x-\sin x}{\cos x+\sin x}dx$

$=\int \frac{dt}{t}$

$=\log \left|t\right|$

$=\log |\cos x+\sin x|+c$ where $c$ is the constant of integration.