Question

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

Answer 1

$$\begin{gathered} \int {\left(a\tan x+b\cot x\right)}^{2}dx \\ =\int \left(a^2{\tan }^{2}x+b^2{\cot }^{2}x+2ab\tan x\cot x\right)dx \\ =a^2\int {\tan }^{2}xdx+b^2\int {\cot }^{2}xdx+2ab\int dx \\ =a^2\int \left({\sec }^{2}x-1\right)dx+b^2\int \left({\mathrm{cosec}}^{2}x-1\right)dx+2ab\int dx \\ =a^2\left[\tan x-x\right]+b^2\left[-\cot x-x\right]+2abx+\mathrm{C} \\ =a^2\tan x-b^2\cot x-\left(a^2+b^2-2ab\right)x+C \end{gathered}$$