Evaluate the following integrals :\(\int\frac{(sin^{-1}x)^3}{\sqrt{1-x

1.	Evaluate the following integrals :\(\int\frac{(sin^{-1}x)^3}{\sqrt{1-x^2}}\) dx
Answer» Assume, sin^-1x = t d( sin^-1x) = dt ⇒ \(\frac{dx}{\sqrt{1-x^2}}\) = dt ∴ Substituting t and dt in given equation we get ⇒ \(\int t^3\) dt ⇒ \(\frac{t^4}{4}\) + c But, t = sin^-1x ⇒ \(\frac{(sin^{-1}x)^4}{4}\) + c

Evaluate the following integrals :\(\int\frac{(sin^{-1}x)^3}{\sqrt{1-x^2}}\) dx

Answer»

Assume,

sin^-1x = t

d( sin^-1x) = dt

⇒ \(\frac{dx}{\sqrt{1-x^2}}\) = dt

∴ Substituting t and dt in given equation we get

⇒ \(\int t^3\) dt

⇒ \(\frac{t^4}{4}\) + c

But,

t = sin^-1x

⇒ \(\frac{(sin^{-1}x)^4}{4}\) + c