Question

The value of the $\int_{{}}^{{}}{\frac{\sin x+\cos x}{3+\sin 2x}}dx$ is

• A. $\frac{1}{4}\ln \left( \frac{2-\sin x+\cos x}{2+\sin x+\cos x} \right)+c$
• B. $\frac{1}{2}\ln$ $\left( \frac{2+\sin x}{2-\sin x} \right)+c$
• C. $\frac{1}{4}\ln$ $\left( \frac{1+\sin x}{1-\sin x} \right)+c$
• D. none of the above

Answer 1

Answer: (A)

$\frac{1}{4}\ln \left( \frac{2-\sin x+\cos x}{2+\sin x+\cos x} \right)+c$

Answer 2

Let $I=\int \frac{\sin x+\cos x}{3+\sin 2 x} d x$
$=\int \frac{\sin x+\cos x}{4+\sin 2 x-1} d x$
$\Rightarrow I=\int \frac{\sin x+\cos x}{4-(\sin x-\cos x)^{2}} d x$
Let $\sin x-\cos x=t$
$(\cos x+\sin x) d x=d t$
$\therefore I=\int \frac{d t}{4-t^{2}}=\frac{1}{4} \log \frac{2-t}{2+t}+c$
$=\frac{1}{4} \log \frac{2-(\sin x-\cos x)}{2+(\sin x+\cos x)}+c$
$=\frac{1}{4} \log \left(\frac{2-\sin x+\cos x}{2+\sin x+\cos x}\right)+c$