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

I = \int {{{\left( {{{\sin }^{ - 1}}x} \right)}^2}dx} \,\,\,\,{\text{ - - - }}\left( {\text{i}} \right)

Put 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 as first and second functions and integration by parts, we have.

I = \int {{z^2}\cos zdz} \,\,\,\,{\text{ - - - }}\left( {{\text{ii}}} \right)

Using 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} } }

Equation (ii) becomes using above formula, we have

\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, we get

 \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