Question

Q1.what is integration of 2 tan x upon 1 + tan squared x dx.

Answer 1

Step-by-step explanation:

Answer:

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

Assuming ( 1 / sec²x ) = t, we get:

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

Hence we get:

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

This is the required answer.