Question

$∫\frac{cos2x}{(sinx+cosx)^2}dx$ is equal to

• A. $\frac{-1}{sinx+cosx}+C$
• B. $log|sinx+cosx|+C$
• C. $log|sinx-cosx|+C$
• D. $\frac{1}{(sinx+cosx)^2}$

Answer 1

Answer: (B)

$log|sinx+cosx|+C$

Answer 2

The correct answer is B:$=log|cosx+sinx|+C$
Let $I=∫\frac{cos2x}{(sinx+cosx)^2}$
$I=∫\frac{cos^2x-sin^2x}{(cosx+sinx)}dx$
$=∫\frac{(cosx+sinx)(cosx-sinx)}{(cosx+sinx)^2}dx$
$=∫\frac{cosx-sinx}{cosx+sinx}dx$
Let $cosx+sinx=t⇒(cosx-sinx)dx=dt$
$∴I=∫\frac{dt}{t}$
$=log|t|+C$
$=log|cosx+sinx|+C$
Hence,the correct Answer is B.