Here you will learn proof of integration of cos inverse x or arccos x and examples based on it.
Letโs begin โ
Integration of Cos Inverse x
The integration of cos inverse x or arccos x is \(xcos^{-1}x\) โ \(\sqrt{1 โ x^2}\) + C
Where C is the integration constant.
i.e. \(\int\) \(cos^{-1}x\) = \(xcos^{-1}x\) โ \(\sqrt{1 โ x^2}\) + C
Proof :ย
We have, I = \(\int\) \(cos^{-1}x\) dx
Let \(cos^{-1}x\) = t,
Then, x = cos t
\(\implies\) dx = d(cos t) = -sin t dt
\(\therefore\) I = \(\int\) \(cos^{-1}x\) dx
\(\implies\) I = \(\int\) -t sint dt
By using integration by parts formula,
I = t cos t + \(\int\) 1. (-cos t) dt
I = t cos t โ \(\int\) cost dt
= t cos t โ sin t + C
= t cos t โ \(\sqrt{1 โ cos^2 t}\) + C
Now, Put t = \(cos^{-1}x\) here
\(\implies\) I = x \(cos^{-1}x\) โ \(\sqrt{1 โ x^2}\) + C
Hence, \(\int\) \(cos^{-1}x\) dx = x \(cos^{-1}x\) โ \(\sqrt{1 โ x^2}\) + C
Example : Evaluate \(\int\) \(x cos^{-1} x\) dx
Solution : We have,
I = \(\int\)ย \(x cos^{-1} x\) dx
By using integration by parts formula,
I = \(cos^{-1} x\) \(x^2\over 2\) โ \(\int\) \(-1\over \sqrt{1 โ x^2}\) \(\times\) \(x^2\over 2\) dx
I =ย \(x^2\over 2\) \(cos^{-1} x\) โ \(1\over 2\) \(\int\) \(-x^2\over \sqrt{1 โ x^2}\) dx
= \(x^2\over 2\) \(cos^{-1} x\) โ \(1\over 2\) \(\int\) \(1 โ x^2 โ 1\over \sqrt{1 โ x^2}\)ย dx
= \(x^2\over 2\) \(cos^{-1} x\) โ \(1\over 2\) { \(\int\) \(1 โ x^2\over \sqrt{1 โ x^2}\) โ \(\int\) \(1\over \sqrt{1 -x^2}\) } dx
\(\implies\) I = \(x^2\over 2\) \(cos^{-1} x\) โ \(1\over 2\) { \(\int\) \(\sqrt{1 โ x^2}\) โ \(\int\) \(1\over \sqrt{1 -x^2}\) } dx
By using integration formula of \(\sqrt{a^2 โ x^2}\),
\(\implies\) I = \(x^2\over 2\) \(cos^{-1} x\) โ \(1\over 2\) [{ \(1\over 2\) \(x\sqrt{1 โ x^2}\) โ \(1\over 2\) \(sin^{-1} x\) } โ \(sin ^{-1} x\) ] + C
\(\implies\) I = \(x^2\over 2\) \(cos^{-1} x\) โย \(1\over 4\) \(x\sqrt{1 โ x^2}\) + \(3\over 4\) \(sin^{-1} x\) + C
Related Questions
What is the Differentiation of cos inverse x ?
