Question

Integrate the given function $x{{\sin }^{-1}}x$?

Answer 1

We have multiplication of two functions involved in $x{{\sin }^{-1}}x$. To integrate $x{{\sin }^{-1}}x$, we use integration by parts by taking ${{\sin }^{-1}}x$ as the function outside of the integral. Once we have done the by parts, we make the changes in the denominator to get the value of all the integrals involved in this calculation. We make the necessary reductions to get the required result.

Complete step-by-step solution:
Given that we have function $x{{\sin }^{-1}}x$ and we need to integrate that given function.
So, we need to find the value of $\int{x{{\sin }^{-1}}xdx}$.
To solve the problem, we need to use the result of integration by parts. We know that integration by parts is to be done when a combination of two or more functions are multiplied to each other inside the integral.
We know that integration of the function fg is defined as $\int{fgdx=f\int{gdx}-\int{\left( \left( \dfrac{df}{dx} \right).\int{gdx} \right)}dx}$.
Let us take $f\left( x \right)={{\sin }^{-1}}\left( x \right)$ and g(x) = x.
We have got $\int{x{{\sin }^{-1}}xdx}=\left( {{\sin }^{-1}}x\int{xdx} \right)-\int{\left( \left( \dfrac{d\left( {{\sin }^{-1}}x \right)}{dx} \right).\int{xdx} \right)}dx$.
We know that $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C}$ and $\dfrac{d\left( {{\sin }^{-1}}x \right)}{dx}=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$.
We have got $\int{x{{\sin }^{-1}}xdx}=\left( {{\sin }^{-1}}\left( x \right).\dfrac{{{x}^{2}}}{2} \right)-\int{\left( \dfrac{1}{\sqrt{1-{{x}^{2}}}}.\dfrac{{{x}^{2}}}{2} \right)}dx$.
We have got $\int{x{{\sin }^{-1}}xdx}=\left( \dfrac{{{x}^{2}}.{{\sin }^{-1}}x}{2} \right)-\dfrac{1}{2}\times \left( \int{\dfrac{{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}dx} \right)$.
We have got $\int{x{{\sin }^{-1}}xdx}=\left( \dfrac{{{x}^{2}}.{{\sin }^{-1}}x}{2} \right)+\dfrac{1}{2}\times \left( \int{\dfrac{1-{{x}^{2}}-1}{\sqrt{1-{{x}^{2}}}}dx} \right)$.
We know that $\int{\left( a+b \right)dx=\int{adx+\int{bdx}}}$.
We have got $\int{x{{\sin }^{-1}}xdx}=\left( \dfrac{{{x}^{2}}.{{\sin }^{-1}}x}{2} \right)+\dfrac{1}{2}\times \left( \int{\dfrac{1-{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}dx} \right)-\dfrac{1}{2}\times \left( \int{\dfrac{1}{\sqrt{1-{{x}^{2}}}}dx} \right)$.
We have $\int{x{{\sin }^{-1}}xdx}=\left( \dfrac{{{x}^{2}}.{{\sin }^{-1}}x}{2} \right)+\dfrac{1}{2}\times \left( \int{\sqrt{1-{{x}^{2}}}dx} \right)-\dfrac{1}{2}\times \left( \int{\dfrac{1}{\sqrt{1-{{x}^{2}}}}dx} \right)$.
We know that $\int{\sqrt{1-{{x}^{2}}}dx=\dfrac{x\times \sqrt{1-{{x}^{2}}}}{2}+\dfrac{1}{2}{{\sin }^{-1}}x+C}$ and $\int{\dfrac{1}{\sqrt{1-{{x}^{2}}}}dx={{\sin }^{-1}}x}+C$.
So, we have $\int{x{{\sin }^{-1}}xdx}=\left( \dfrac{{{x}^{2}}.{{\sin }^{-1}}x}{2} \right)+\dfrac{1}{2}\times \left( \dfrac{x\times \sqrt{1-{{x}^{2}}}}{2}+\dfrac{1}{2}{{\sin }^{-1}}x \right)-\dfrac{1}{2}\times \left( {{\sin }^{-1}}x \right)+C$.
We have $\int{x{{\sin }^{-1}}xdx}=\dfrac{{{x}^{2}}.{{\sin }^{-1}}x}{2}+\dfrac{x\times \sqrt{1-{{x}^{2}}}}{4}+\dfrac{1}{4}{{\sin }^{-1}}x-\dfrac{1}{2}{{\sin }^{-1}}x+C$.
We have $\int{x{{\sin }^{-1}}xdx}=\dfrac{{{x}^{2}}.{{\sin }^{-1}}x}{2}+\dfrac{x\times \sqrt{1-{{x}^{2}}}}{4}-\dfrac{1}{4}{{\sin }^{-1}}x+C$.
We have $\int{x{{\sin }^{-1}}xdx}=\dfrac{x\times \sqrt{1-{{x}^{2}}}}{4}+\dfrac{{{\sin }^{-1}}x\left( 2{{x}^{2}}-1 \right)}{4}+C$.
We got the value of $\int{x{{\sin }^{-1}}xdx}$ as $\dfrac{x\times \sqrt{1-{{x}^{2}}}}{4}+\dfrac{{{\sin }^{-1}}x\left( 2{{x}^{2}}-1 \right)}{4}+C$.
$\therefore$ The integral $\int{x{{\sin }^{-1}}xdx}=\dfrac{x\times \sqrt{1-{{x}^{2}}}}{4}+\dfrac{{{\sin }^{-1}}x\left( 2{{x}^{2}}-1 \right)}{4}+C$.

Note: Whenever we see the integration problem involving inverse trigonometric functions, we use integrations by parts to get the integration. We always take inverse functions as ‘f’ for doing by parts as they cannot be integrated directly. While calculating the integral by parts, we need to make sure that no signs are gone wrong.