what is the answer to : integration of tan-1 (secx+tanx) dx - Maths - Integrals

Question

what is the answer to :
integration of tan-1 (secx+tanx) dx ??

Ajanta Trivedi · Accepted Answer

to find the given integration, let us take

$\tan^{- 1} (\sec x + \tan x) = t \frac{1}{1 + (\sec x + \tan x)^{2}} (\sec x \tan x + \sec^{2} x) d x = d t \frac{1}{1 + \sec^{2} x + \tan^{2} x + 2 \sec x \tan x} (\sec x \tan x + \sec^{2} x) d x = d t \frac{1}{\sec^{2} x + \sec^{2} x + 2 \sec x \tan x} (\sec x \tan x + \sec^{2} x) d x = d t [\sin c e 1 + \tan^{2} x = \sec^{2} x] \frac{1}{2 (\sec^{2} x + \sec x \tan x)} (\sec x \tan x + \sec^{2} x) d x = d t \frac{1}{2} d x = d t \Rightarrow d x = 2 d t$

therefore the given integration can be written as

$\int \tan^{- 1} (\sec x + \tan x) ⅆ x = \int t \cdot 2 ⅆ t = \frac{t^{2}}{2} \cdot 2 + C = t^{2} + C$ where C is an integration constant.

therefore $\int \tan^{- 1} (\sec x + \tan x) ⅆ x = {[\tan^{- 1} (\sec x + \tan x)]}^{2} + C$

what is the answer to : integration of tan-1 (secx+tanx) dx ??

what is the answer to :

integration of tan^-1 (secx+tanx) dx ??