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


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 ?

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