Question

Find the following integral
$\left(i\right)\int (\sin x+\cos x)dx$
$\left(ii\right)\int \text{cosec}x(\text{cosec}x+\cot x)dx$
$\left(iii\right)\int \frac{1-\sin x}{{\cos }^{2}x}dx$

Answer 1

$\left(i\right)\int (\sin x+\cos x)dx$
$=\int \left(\sin x\right)dx+\int \left(\cos x\right)dx$
$=-\cos x+\sin x+C$

Where $C$ is constant of integration

$\left(ii\right)\int \text{cosec}x(\text{cosec}x+\cot x)dx$
$=\int ({\text{cosec}}^{2}x+\text{cosec}x\cot x)dx$
$=\int {\text{cosec}}^{2}xdx+\int \text{cosec}x\cot xdx$
$=-\cot x-\text{cosec}x+C$

Where $C$ is constant of integration

$\left(iii\right)\int \frac{1-\sin x}{{\cos }^{2}x}dx$
$=\int (\frac{1}{{\cos }^{2}x}-\frac{\sin x}{{\cos }^{2}x})dx$
$=\int ({\sec }^{2}x-\frac{\sin x}{\cos x}\frac{1}{\cos x})dx$
$=\int ({\sec }^{2}x-\tan x\sec x)dx$
$=\int {\sec }^{2}xdx-\int (\tan x.\sec x)dx$
$=\tan x-\sec x+C$

Where $C$ is constant of integration