# Integral of Inverse Sine Squared

In this tutorial we shall find the integral of inverse sine squared function, and it is another important integration. To evaluate this integral we first use the method of substitution and then we use integration by parts.

The integral of inverse sine squared is of the form
$I = \int {{{\left( {{{\sin }^{ – 1}}x} \right)}^2}dx} \,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

But $z = {\sin ^{ – 1}}x$ implies that $\sin z = x$, by differentiation $\cos zdz = dx$, so the given integral (i) takes the form
$I = \int {{{\left( {{{\sin }^{ – 1}}x} \right)}^2}dx} = \int {{z^2}\cos zdz}$

Considering ${z^2}$ and $\cos z$ are the first and second functions and using integration by parts, we have
$I = \int {{z^2}\cos zdz} \,\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right)$

Using the formula for integration by parts, we have
$\int {\left[ {f\left( x \right)g\left( x \right)} \right]dx = f\left( x \right)\int {g\left( x \right)dx – \int {\left[ {\frac{d}{{dx}}f\left( x \right)\int {g\left( x \right)dx} } \right]dx} } }$

Using the formula above, equation (ii) becomes
$\begin{gathered} I = {z^2}\int {\cos zdz – \int {\left[ {\frac{d}{{dz}}{z^2}\int {\cos zdz} } \right]dz} } \\ \Rightarrow I = {z^2}\sin z – 2\int {z\sin zdz} \\ \end{gathered}$

Now again using integration by parts, we have
$\begin{gathered} I = {z^2}\sin z – 2\left[ {z\int {\sin zdz – \int {\left( {\frac{d}{{dz}}z\int {\sin zdz} } \right)dz} } } \right] \\ \Rightarrow I = {z^2}\sin z – 2\left[ {z\left( { – \cos z} \right) – \int {\left( { – \cos z} \right)dz} } \right] \\ \Rightarrow I = {z^2}\sin z – 2\left[ { – z\cos z + \int {\cos zdz} } \right] \\ \Rightarrow I = {z^2}\sin z + 2z\cos z – 2\sin z + c \\ \Rightarrow I = {z^2}\sin z + 2z\sqrt {1 – {{\sin }^2}z} – 2\sin z + c \\ \end{gathered}$

From the above substitution it can be written in the form
$\Rightarrow \int {{{\left( {{{\sin }^{ – 1}}x} \right)}^2}} = x{\left( {{{\sin }^{ – 1}}x} \right)^2} + 2\left( {{{\sin }^{ – 1}}x} \right)\sqrt {1 – {x^2}} – 2x + c$