Integration of Tanx

Here you will learn proof of integration of tanx or tan x and examples based on it.

Let’s begin –

Integration of Tanx or Tan x

The integration of tanx is – log |cos x| + C or log |sec x| + C

i.e. $\int$ (tanx) dx = – log |cos x| + C or,

$\int$ (tanx) dx = log |sec x| + C

Proof :  

Let I = $\int$ (tan x) dx

Then, I = $\int$ $sin x\over cos x$ dx

Let cos x = t 

Then, d(cos x) = dt $\implies$ -sin x dx = dt 

$\implies$ dx = $-dt\over sin x$

Putting cos x = t, and dx = $-dt\over sin x$, we get

I = $\int$ $sin x\over cos x$ $\times$ $-dt\over sin x$

= $\int$ $-1\over t$ dt = – log |t| + C

= – log |cos x| + C

And cos x = $1\over sec x$

$\implies$ I = -log |1/sec x| + C = -$log |sec^{-1} x|$ + C = log |sec x| + C

Hence, $\int$ (tanx) dx = – log |cos x| + C or, $\int$ (tanx) dx = log |sec x| + C

Example : Evaluate : $\int$ $\sqrt{{1-cos 2x}\over {1+cos 2x}}$ dx

Solution : We have, 

I = $\int$ $\sqrt{{1-cos 2x}\over {1+cos 2x}}$ dx

By Trigonometry formulas,

1 – cos 2x = $2sin^2 x$ and 1 + cos 2x = $2cos^2 x$

$\implies$ I = $\int$ $\sqrt{{2sin^2 x}\over {2cos^2 x}}$ dx

$\implies$ I = $\int$ ${sin x}\over {cos x}$ dx

{$\because$ ${sin x}\over {cos x}$ = tan x }

$\implies$ I = $\int$ tan x dx                       

$\implies$ I = log |sec x| + C = – log |cos x| + C

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Related Questions

What is the Differentiation of tan x ?

What is the Integration of tan inverse x ?

What is the Differentiation of tan inverse x ?