Question

Evaluate
(tan2x + cot2x)dx.​

Answer 1

Given Integrand,

$\displaystyle \sf \int (tan2x + cot2x)dx$

Consider t = 2x.

Differentiating both sides w.r.t x,

$\longrightarrow$ dt/dx = 2

$\longrightarrow$ dx = dt/2.

Accordingly,

$\longrightarrow\displaystyle \sf \dfrac{1}{2} \int \{tan(t)+ cot(t) \}dt \\ \\ \longrightarrow\displaystyle \sf \dfrac{1}{2} \int tan(t)dt + \dfrac{1}{2} \int cot(t)dt \\ \\ \longrightarrow\displaystyle \sf \dfrac{1}{2} ln(sec \: 2x) + \dfrac{1}{2} ln(sin \: 2x) + c \\ \\ \longrightarrow\displaystyle \sf \dfrac{1}{2} ln(sec \: 2x.sin \: 2x) + c \\ \\ \longrightarrow\displaystyle \boxed{ \boxed{\sf ln( \sqrt{tan2x} ) + c}}$

Answer 2

Please do go through the attachment for the explanation.