Question

Evaluate:

$\int {\left(a\tan x+b\cot x\right)}^{2}dx$

Answer 1

$I=\int (a\tan x+b\cot x{)}^{2}dx$

$=\int (a^2{\tan }^{2}x+b^2{\cot }^{2}x+2ab)dx$

$I=\int a^2({\sec }^{2}x-1)+b^2(cose{c}^{2}x-1)+2ab)dx$

$I=\int a^2{\sec }^{2}xdx+\int b^{2}cose{c}^{2}dx-\int (a-b{)}^{2}dx$

$I=a^2\tan x+\int b^{2}cose{c}^{2}xdx-\int (a-b{)}^{2}dx$

$I=a^2\tan x-b^2\cot x-\int (a-b{)}^{2}dx$

$I=a^2\tan x-b^2\cot x-(a-b{)}^{2}x+c$.