Here you will learn proof of integration of cotx or cot x and examples based on it.
Let’s begin –
Integration of Cotx or Cot x
The integration of cotx is log |sin x| + C or – log |cosec x| + C
i.e. \(\int\) (cotx) dx = log |sin x| + C or,
\(\int\) (cotx) dx = -log |cosec x| + C
Proof :
Let I = \(\int\) (cot x) dx
Then, I = \(\int\) \(cos x\over sin x\) dx
Let sin x = t
Then, d(sin x) = dt \(\implies\) cos x dx = dt
\(\implies\) dx = \(dt\over cos x\)
Putting sin x = t, and dx = \(dt\over cos x\), we get
I = \(\int\) \(cos x\over sin x\) \(\times\) \(dt\over cos x\)
= \(\int\) \(1\over t\) dt = log |t| + C
= log |sin x| + C
And sin x = \(1\over cosec x\)
\(\implies\) I = log |1/cosec x| + C = \(log |cosec^{-1} x|\) + C = -log |cosec x| + C
Hence, \(\int\) (cotx) dx = log |sin x| + C or, \(\int\) (cotx) dx = -log |cosec 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 = \(\int\) \(\sqrt{{2cos^2 x}\over {2sin^2 x}}\) dx
\(\implies\) I = \(\int\) \({cos x}\over {sin x}\) dx
{\(\because\) \({cos x}\over {sin x}\) = cot x }
\(\implies\) I = \(\int\) cot x dx
\(\implies\) I = log |sin x| + C = – log |cosec x| + C
Related Questions
What is the Differentiation of cot x ?
