# Integral of Inverse Sine Squared

In this tutorial we shall find integral of inverse sine squared function, it is also an important integration, to evaluate this integral first use method of substitution then use integration by parts.

The integral of inverse sine squared is of the form

Put $z = {\sin ^{ - 1}}x$ implies that $\sin z = x$, by differentiation $\cos zdz = dx$, so the given integral (i) takes the form

Considering ${z^2}$ and $\cos z$ as first and second functions and integration by parts, we have.

Using formula for integration by parts, we have

Equation (ii) becomes using above formula, we have

Now again using integration by parts, we have

From the above substitution it can be written in the form, we get