Question

what is integration of 2 tan x upon 1 + tan squared x dx

Answer 1

it can be written as tan2x by applying the formula of tan2x
which means we have to integrate tan2x
= sec^2x × (2x^2)/2
= sec^2x × x^2
= (x × sec2x)^2.

Answer 2

Answer:

$\implies \int {\dfrac{2.tan\:x}{1+tan^2\:x}} \, dx \\\\\\\implies \int {\dfrac{2.tan\:x}{sec^2\:x}} \, dx$

Assuming ( 1 / sec²x ) = t, we get:

$\implies \dfrac{dt}{dx} = \dfrac{d}{dx}\:(\dfrac{1}{sec^2\:x})\\\\\\\implies \dfrac{dt}{dx} = \dfrac{-2.\:tan\;x}{sec^2\:x}\\\\\\\implies dt = \dfrac{-2.\:tan\;x}{sec^2\:x}.dx$

Hence we get:

$\implies \int {-1} \, dt \\\\\\\implies \boxed{ \bf{ f(x) = -t + C}} \\\\\\\text{Substituting the values we get:}\\\\\\\implies \boxed{ \bf{ f(x) = \dfrac{-1}{sec^2\:x} + C}}$

This is the required answer.