Question

The value of the integral $\int \frac{cosx}{sinx+cosx}dx$ is equal to

• A. $x+log|sinx+cosx|+C$
• B. $\frac{1}{2}\left[x+log\left|\begin{array}{c}sinx+cosx\end{array}\right|\right]+C$
• C. $log|sinx+cosx|+C$
• D. $\frac{x}{2}+log\left|\begin{array}{c}sinx+cosx\end{array}\right|+C$
• E. $x+\frac{1}{2}+log\left|\begin{array}{c}sinx+cosx\end{array}\right|+C$

Answer 1

Answer: (B)

Let $I=\int \frac{\cos x}{\sin x+\cos x}dx$
$=\frac{1}{2}\int \frac{2\cos x}{\sin x+\cos x}dx$
$=\frac{1}{2}\int \frac{(\sin x+\cos x)+(\cos x-\sin x)}{(\sin x+\cos x)}dx$
$=\frac{1}{2}\int dx+\frac{1}{2}\int \frac{\cos x-\sin x}{\sin x+\cos x}\cdot dx$
$=\frac{1}{2}x+\frac{1}{2}\log |\sin x+\cos x|+C$
$=\frac{1}{2}[x+\log |\sin x+\cos x|]+C$