Question

$\int \frac{sinx+cosx}{sin2x}.dx$

Answer 1

$\int \frac{\sin x+\cos x}{\sin 2x}dx$

Let $\sin x-\cos x=t$

$\Rightarrow dt=(\cos x+\sin x)dx$

$t^2=(\sin x-\cos x{)}^2=1-2\sin x\cos x$

$\Rightarrow t^2=1-\sin 2x$

$\Rightarrow \sin 2x=1-t^2$

(split into partial fractions)

$\int \frac{(\sin x+\cos x)dx}{\sin 2x}=\int \frac{dt}{1-t^2}=\frac{1}{2}\int \frac{(1+t)+(1-t)}{(1-t)(1+t)}dt$

$=\frac{1}{2}\left[\int \frac{1}{1-t}dt+\int \frac{1}{1+t}dt\right]$

$=\frac{1}{2}[-ln|1-t|+ln|1+t|]+c$

$=\frac{1}{2}ln\left|\frac{1+t}{1-t}\right|+c$

$=\frac{1}{2}ln\left|\frac{1+\sin x-\cos x}{1-\sin x+\cos x}\right|+c$.