Question

If $\int \frac{1}{1+\tan x}dx=Ax+B\ln |\cos x+\sin x|+C,$ then the value of ${(A+B)}^5$ is equal to
(where $C$ is integration constant)

Answer 1

$I=\int \frac{\cos x}{\cos +\sin x}dx=\frac{1}{2}\int \frac{2\cos x}{\cos +\sin x}dx=\frac{1}{2}\int \frac{(\cos x+\sin x)+(\cos x-\sin x)}{\cos +\sin x}dx=\frac{1}{2}\int dx+\frac{1}{2}\int \frac{\cos x-\sin x}{\cos x+\sin x}dx=\frac{x}{2}+\frac{1}{2}\ln |\cos x+\sin x|+C\Rightarrow A=\frac{1}{2},B=\frac{1}{2}\Rightarrow (A+B{)}^5=1$