Question

integral of secx / ( secx + tanx) ^2. dx

Answer 1

what is your question

Answer 2

The answer is  $\frac{-1}{(sec x + tanx)^2}$ + c

Given,

$\frac{secx}{( secx + tanx )^2}$

To Find,

Integration of  $\frac{secx}{( secx + tanx )^2}$

Solution,

Let $\frac{1}{( secx + tanx )^2}$ = t

Differentiate on both the sides

⇒ $\frac{-1}{( secx + tanx )^3}$ × (sec x tan x + sec² x) = $\frac{dx}{dt}$

⇒ $\frac{- secx ( secx + tanx )}{( secx + tanx)^3}$ = $\frac{-dt}{dx}$

⇒ $\frac{secx dx}{( secx + tan )^2}$ = -dt

∴ ∫ - dt

∴ -t + c

Hence, the answer is $\frac{-1}{(sec x + tanx)^2}$ + c