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 ?
