Here you will learn proof of integration of sin inverse x or arcsin x and examples based on it.
Let’s begin –
Integration of Sin Inverse x
The integration of sin inverse x or arcsin x is \(xsin^{-1}x\) + \(\sqrt{1 – x^2}\) + C
Where C is the integration constant.
i.e. \(\int\) \(sin^{-1}x\) = \(xsin^{-1}x\) + \(\sqrt{1 – x^2}\) + C
Proof :
We have, I = \(\int\) \(sin^{-1}x\) dx
Let \(sin^{-1}x\) = t,
Then, x = sint
\(\implies\) dx = d(sin t) = cos t dt
\(\therefore\) I = \(\int\) \(sin^{-1}x\) dx
\(\implies\) I = \(\int\) t cost dt
By using integration by parts formula,
I = t sin t – \(\int\) 1. (sin t) dt
I = t sin t – \(\int\) sint dt
= t sin t + cos t + C
= t sin t + \(\sqrt{1 – sin^2 t}\) + C
Now, Put t = \(sin^{-1}x\) here
\(\implies\) I = x \(sin^{-1}x\) + \(\sqrt{1 – x^2}\) + C
Hence, \(\int\) \(sin^{-1}x\) dx = x \(sin^{-1}x\) + \(\sqrt{1 – x^2}\) + C
Example : Evaluate \(\int\) \(x sin^{-1} x\) dx
Solution : We have,
I = \(\int\) \(x sin^{-1} x\) dx
By using integration by parts formula,
I = \(sin^{-1} x\) \(x^2\over 2\) – \(\int\) \(1\over \sqrt{1 – x^2}\) \(\times\) \(x^2\over 2\) dx
I = \(x^2\over 2\) \(sin^{-1} x\) + \(1\over 2\) \(\int\) \(-x^2\over \sqrt{1 – x^2}\) dx
= \(x^2\over 2\) \(sin^{-1} x\) + \(1\over 2\) \(\int\) \(1 – x^2 – 1\over \sqrt{1 – x^2}\) dx
= \(x^2\over 2\) \(sin^{-1} x\) + \(1\over 2\) { \(\int\) \(1 – x^2\over \sqrt{1 – x^2}\) – \(\int\) \(1\over \sqrt{1 -x^2}\) } dx
\(\implies\) I = \(x^2\over 2\) \(sin^{-1} x\) + \(1\over 2\) { \(\int\) \(\sqrt{1 – x^2}\) – \(\int\) \(1\over \sqrt{1 -x^2}\) } dx
By using integration formula of \(\sqrt{a^2 – x^2}\),
\(\implies\) I = \(x^2\over 2\) \(sin^{-1} x\) + \(1\over 2\) [{ \(1\over 2\) \(x\sqrt{1 – x^2}\) – \(1\over 2\) \(sin^{-1} x\) } – \(sin^{-1} x\) ] + C
\(\implies\) I = \(x^2\over 2\) \(sin^{-1} x\) + \(1\over 4\) \(x\sqrt{1 – x^2}\) – \(1\over 4\) \(sin^{-1} x\) + C
Related Questions
What is the Differentiation of sin inverse x ?