Question

Evaluate: $\int \frac{1}{1-\sin x}dx$.

Answer 1

$=\int \frac{1}{1-\sin x}dx$

$=\int \frac{1}{1-\sin x}\times \left(\frac{1+\sin x}{1+\sin x}\right)dx$

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

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

$\Rightarrow =\int \left(\frac{1}{{\cos }^{2}x}+\frac{\sin x}{{\cos }^{2}x}\right)dx$

$=\int {\sec }^{2}xdx+\int \sec x\tan xdx$

$\Rightarrow =\tan x+\sec x+c$.