Integration of x Sin Inverse x

In this tutorial we shall find the integral of x sine inverse of x, and solve this problem with the help of the integration by parts methods

The integral of x sine inverse of x is of the form

I = \int {x{{\sin }^{ - 1}}xdx}

Here the first function is {\sin ^{ - 1}}x and the second function is x.

I = \int {{{\sin }^{ - 1}}x \cdot xdx} \,\,\,\,{\text{ - - - }}\left( {\text{i}} \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 (i) becomes

\begin{gathered} I = {\sin ^{ - 1}}x\int {xdx - \int {\left[ {\frac{d}{{dx}}{{\sin }^{ - 1}}x\left( {\int {xdx} } \right)} \right]} dx} \\ \Rightarrow I = {\sin ^{ - 1}}x\frac{{{x^2}}}{2} - \int {\left[ {\frac{1}{{\sqrt {1 - {x^2}} }}\frac{{{x^2}}}{2}} \right]} dx \\ \Rightarrow I = \frac{{{x^2}}}{2}{\sin ^{ - 1}}x - \frac{1}{2}\int {\frac{{{x^2}}}{{\sqrt {1 - {x^2}} }}} dx \\ \Rightarrow I = \frac{{{x^2}}}{2}{\sin ^{ - 1}}x + \frac{1}{2}\int {\frac{{1 - {x^2} - 1}}{{\sqrt {1 - {x^2}} }}} dx \\ \Rightarrow I = \frac{{{x^2}}}{2}{\sin ^{ - 1}}x + \frac{1}{2}\int {\frac{{1 - {x^2}}}{{\sqrt {1 - {x^2}} }}} dx - \frac{1}{2}\int {\frac{1}{{\sqrt {1 - {x^2}} }}dx} \\ \Rightarrow I = \frac{{{x^2}}}{2}{\sin ^{ - 1}}x + \frac{1}{2}\int {\sqrt {1 - {x^2}} } dx - \frac{1}{2}\int {\frac{1}{{\sqrt {1 - {x^2}} }}dx} \\ \end{gathered}

Using the integral formula \int {\sqrt {{a^2} - {x^2}} } dx = \frac{{x\sqrt {{a^2} - {x^2}} }}{2} + \frac{{{a^2}}}{2}{\sin ^{ - 1}}\left( {\frac{x}{a}} \right) + c, we have

\begin{gathered} I = \frac{{{x^2}}}{2}{\sin ^{ - 1}}x + \frac{1}{2}\left[ {\frac{{x\sqrt {1 - {x^2}} }}{2} - \frac{1}{2}{{\sin }^{ - 1}}x} \right] - \frac{1}{2}{\sin ^{ - 1}}x + c \\ \Rightarrow I = \frac{{{x^2}}}{2}{\sin ^{ - 1}}x - \frac{3}{2}{\sin ^{ - 1}}x + \frac{{x\sqrt {1 - {x^2}} }}{4} + c \\ \end{gathered}