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