x x+ x d x = x x+ x x- x x- x d x = x x- ^2 x ^2 x- ^2 x d x = x x- ( ^2 x-1 ) 1 d x = x x d x- ^2 x d x+ 1 d x = x- x+x+c

Evaluate: x x+ x d x

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Question

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

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Answer

$ \int \frac{\tan x}{\sec x+\tan x} d x$
$=\int \frac{\tan x}{\sec x+\tan x} \times \frac{\sec x-\tan x}{\sec x-\tan x} d x$
$=\int \frac{\sec x \tan x-\tan ^2 x}{\sec ^2 x-\tan ^2 x} d x$
$=\int \frac{\sec x \tan x-\left(\sec ^2 x-1\right)}{1} d x$ $=\int \sec x \tan x d x-\int \sec ^2 x d x+\int 1 d x$
$=\sec x-\tan x+x+c$

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