I = ∫ x x+tan xdx = ∫ x( x + tan x)( x+tan x)^2dx = ∫(^2 x + xtan x)( x+tan x)^2dx Let t = ( x+tan x) dt = (^2 x + xtan x)dx ∴ I = ∫1t^ ...

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First Principle of Differentiation

Solve: ∫ x ...

Question

Solve: $\int \frac{sec x}{sec x + tan x} d x$

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Solution

I = $\int \frac{sec x}{sec x + tan x} d x$
= $\int \frac{sec x (sec x + tan x)}{(sec x + tan x)^{2}} d x$
= $\int \frac{({sec}^{2} x + sec x tan x)}{(sec x + tan x)^{2}} d x$
Let $t = (sec x + tan x) ⟹ d t = ({sec}^{2} x + sec x tan x) d x$
$∴ I = \int \frac{1}{t^{2}} d t$
$⟹ I = \frac{1}{t} + C$
$⟹ I = \frac{1}{sec x + tan x} + C$

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