Integral of Inverse Cosine Squared

In this tutorial we shall find integral of inverse cosine 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 cosine squared is of the form

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


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

I  =  - \int {{z^2}\sin 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 =  -  \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, we get

  \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

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