Question

$\int {\sec }^{2}x.{\text{cosec}}^{2}xdx=$

• A. $\tan x-\cot x+c$
• B. $\tan x+\cot x+c$
• C. $-\tan x+\cot x+c$
• D. $\sec x\tan x+c$

Answer 1

Answer: (A)

$\tan x-\cot x+c$

$\int {\sec }^{2}x.{\text{cosec}}^{2}xdx$
$=\int \frac{1}{{\cos }^{2}x.{\sin }^{2}x}dx$
$=\int \frac{{\cos }^{2}x+{\sin }^{2}x}{{\cos }^{2}x.{\sin }^{2}x}dx$
$=\int \frac{{\cos }^{2}x}{{\cos }^{2}x{\sin }^{2}x}+\frac{{\sin }^{2}x}{{\cos }^{2}x.{\sin }^{2}x}dx$
$=\int \frac{1}{{\sin }^{2}x}+\frac{1}{{\cos }^{2}x}dx$
$=\int ({\text{cosec}}^{2}x+{\sec }^{2}x)dx$
$=-\cot x+\tan x+c$
$=\tan x-\cot x+c$