Integral of Inverse Cosine Squared

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

The integral of inverse cosine squared is of the form

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

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

I = \int {{{\left( {{{\cos }^{ - 1}}x} \right)}^2}dx} = - \int {{z^2}\sin zdz}

Considering {z^2} and \sin z are the first and second functions and using integration by parts, we have

I = - \int {{z^2}\sin 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 = - \left[ {{z^2}\int {\sin zdz - \int {\left[ {\frac{d}{{dz}}{z^2}\int {\sin zdz} } \right]dz} } } \right] \\ \Rightarrow I = - \left[ {{z^2}\left( { - \cos z} \right) - 2\int {z\left( { - \cos z} \right)dz} } \right] \\ \Rightarrow I = {z^2}\cos z - 2\int {z\cos zdz} \\ \end{gathered}

Now again using integration by parts, we have

\begin{gathered} I = {z^2}\cos z - 2\left[ {z\int {\cos zdz - \int {\left[ {\frac{d}{{dz}}z\int {\cos zdz} } \right]dz} } } \right] \\ \Rightarrow I = {z^2}\cos z - 2\left[ {z\sin z - \int {\sin zdz} } \right] \\ \Rightarrow I = {z^2}\cos z - 2\left[ {z\sin z - \left( { - \cos z} \right)} \right] \\ \Rightarrow I = {z^2}\cos z - 2z\sin z - 2\cos z + c \\ \Rightarrow I = {z^2}\cos z - 2z\sqrt {1 - {{\cos }^2}z} - 2\cos z + c \\ \end{gathered}

From the above substitution it can be written in the form

 \Rightarrow \int {{{\left( {{{\cos }^{ - 1}}x} \right)}^2}} = x{\left( {{{\cos }^{ - 1}}x} \right)^2} - 2\left( {{{\cos }^{ - 1}}x} \right)\sqrt {1 - {x^2}} - 2x + c