1 1 + cos 2 x ? d x =

Question

$\int \frac{1}{1 + \cos^{2} x} d x =$

Smart Achievers · Accepted Answer

(3)

Ans. $\frac{1}{\sqrt{2}} \tan^{- 1} (\frac{1}{\sqrt{2}} \tan x) + C$

Explanation :

$\int \frac{d x}{1 + \cos^{2} x} = \int \frac{\sec^{2} x}{\sec^{2} x + 1} d x = \int \frac{\sec^{2} x}{\tan^{2} x + 2} d x$

$= \int \frac{d t}{t^{2} + 2} = \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{t}{\sqrt{2}}) + C$ ${Putting \tan x = t, \sec^{2} x d x = d t}$

$= \frac{1}{\sqrt{2}} \tan^{- 1} (\frac{1}{\sqrt{2}} \tan x) + C$

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