Question

Integrate the following function with respect to $x$
$\frac{1}{\sin x-\cos x}$

Answer 1

Marks	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$
Frequency	$1$	$2$	$1$	$3$	$5$	$4$	$2$	$1$	$1$

$\int \frac{1}{\sin x-\cos x}\times \frac{\sin x+\cos x}{\sin x+\cos x}$

$\int \frac{\sin x+\cos x}{{\sin }^{2}x-{\cos }^{2}x}\therefore \int \frac{\sin x+\cos x}{2{\sin }^{2}x-1}$

$\Rightarrow -\int \frac{\sin x+\cos x}{1-2{\sin }^{2}x}$

$\Rightarrow \left(1\right)\int \frac{\sin x}{\cos 2x}dx=\frac{1}{2\sqrt{2}}(\ln /2\sin x+\sqrt{2}-\ln /2\sin x-\sqrt{2})$

$\Rightarrow \frac{1}{2\sqrt{2}}\ln |2\sin x+\sqrt{2}|-\ln I2\sin x-\sqrt{2}I$

Thus

$\int \frac{1}{\sin x-\cos x}dx=\int \frac{\cos x-\sin x}{\cos 2x}dx=\int \frac{-\sin x}{\cos 3x}+\int \frac{\cos x}{\cos 3x}dx$