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Evaluate the following indefinite integral: $\int{\dfrac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}dx}$?

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Last updated date: 20th Jun 2024
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Answer
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Hint: We start solving the problem by recalling the uv rule of integration as $\int{u\left( x \right)v\left( x \right)dx}=u\left( x \right)\int{v\left( x \right)dx}-\int{\left( \dfrac{d\left( u\left( x \right) \right)}{dx}\int{v\left( x \right)dx} \right)dx}$ by assuming suitable functions for $u\left( x \right)$ and $v\left( x \right)$. We then use $\dfrac{d\left( {{\sin }^{-1}}x \right)}{dx}=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$ assume $\int{\dfrac{x}{\sqrt{1-{{x}^{2}}}}dx}={{I}_{1}}$. We then solve ${{I}_{1}}$ first by replacing $1-{{x}^{2}}$ with t and $dx$ with $dt$ and using the fact $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C$. We then substitute the value of ${{I}_{1}}$ in I and make necessary calculations to find the answer for the given integral.

Complete step by step answer:
According to the problem, we need to solve the given indefinite integral $\int{\dfrac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}dx}$.
Let us assume the given indefinite integral be $I=\int{\dfrac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}dx}$.
We know that from uv rule of integration, we have $\int{u\left( x \right)v\left( x \right)dx}=u\left( x \right)\int{v\left( x \right)dx}-\int{\left( \dfrac{d\left( u\left( x \right) \right)}{dx}\int{v\left( x \right)dx} \right)dx}$. Let us this rule to solve for I.
So, we have $u\left( x \right)={{\sin }^{-1}}x$ and $v\left( x \right)=\dfrac{x}{\sqrt{1-{{x}^{2}}}}$.
$\Rightarrow I={{\sin }^{-1}}x\int{\dfrac{x}{\sqrt{1-{{x}^{2}}}}dx}-\int{\left( \dfrac{d\left( {{\sin }^{-1}}x \right)}{dx}\int{\dfrac{x}{\sqrt{1-{{x}^{2}}}}dx} \right)}dx$.
We know that $\dfrac{d\left( {{\sin }^{-1}}x \right)}{dx}=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$.
$\Rightarrow I={{\sin }^{-1}}x\int{\dfrac{x}{\sqrt{1-{{x}^{2}}}}dx}-\int{\left( \dfrac{1}{\sqrt{1-{{x}^{2}}}}\int{\dfrac{x}{\sqrt{1-{{x}^{2}}}}dx} \right)}dx$.
Let us assume $\int{\dfrac{x}{\sqrt{1-{{x}^{2}}}}dx}={{I}_{1}}$. So, we get $I={{\sin }^{-1}}x{{I}_{1}}-\int{\left( \dfrac{1}{\sqrt{1-{{x}^{2}}}}{{I}_{1}} \right)}dx$ ---(1).
$\Rightarrow {{I}_{1}}=\int{\dfrac{x}{\sqrt{1-{{x}^{2}}}}dx}$ ---(2).
Let us assume $1-{{x}^{2}}=t$ ---(3). Now, apply differential on both sides.
$\Rightarrow d\left( 1-{{x}^{2}} \right)=d\left( t \right)$.
We know that $d\left( a+b \right)=d\left( a \right)+d\left( b \right)$.
\[\Rightarrow d\left( 1 \right)-d\left( {{x}^{2}} \right)=dt\].
We know that for a constant c $d\left( c \right)=0$ and $d\left( {{x}^{n}} \right)=n{{x}^{n-1}}dx$.
\[\Rightarrow 0-2xdx=dt\].
\[\Rightarrow -2xdx=dt\].
\[\Rightarrow xdx=\dfrac{-dt}{2}\] ---(4).
Let us substitute equation (3) and (4) in equation (2).
So, we get ${{I}_{1}}=\int{\dfrac{-1}{\sqrt{t}}\dfrac{dt}{2}}$.
$\Rightarrow {{I}_{1}}=\dfrac{-1}{2}\int{{{t}^{\dfrac{-1}{2}}}dt}$.
We know that $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C$, where C is constant of integration.
$\Rightarrow {{I}_{1}}=\dfrac{-1}{2}\left( \dfrac{{{t}^{\dfrac{-1}{2}+1}}}{\dfrac{-1}{2}+1} \right)+C$.
$\Rightarrow {{I}_{1}}=\dfrac{-1}{2}\left( \dfrac{{{t}^{\dfrac{1}{2}}}}{\dfrac{1}{2}} \right)+C$.
$\Rightarrow {{I}_{1}}=-{{t}^{\dfrac{1}{2}}}+C$.
From equation (3), we have $1-{{x}^{2}}=t$. So, we have ${{I}_{1}}=-\sqrt{1-{{x}^{2}}}+C$ and let us substitute this in equation (1).
$\Rightarrow I={{\sin }^{-1}}x\left( -\sqrt{1-{{x}^{2}}} \right)-\int{\left( \dfrac{1}{\sqrt{1-{{x}^{2}}}}\left( -\sqrt{1-{{x}^{2}}} \right) \right)}dx$, here we neglected constant of integration as we accommodate in final constant of integration.
$\Rightarrow I=-\sqrt{1-{{x}^{2}}}{{\sin }^{-1}}x-\int{\left( -1 \right)}dx$.
$\Rightarrow I=-\sqrt{1-{{x}^{2}}}{{\sin }^{-1}}x+\int{dx}$.
We know that $\int{adx}=ax+C$.
$\Rightarrow I=-\sqrt{1-{{x}^{2}}}{{\sin }^{-1}}x+x+C$.
So, we have found the result of integration as $\int{\dfrac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}dx}=-\sqrt{1-{{x}^{2}}}{{\sin }^{-1}}x+x+C$.

∴ $\int{\dfrac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}dx}=-\sqrt{1-{{x}^{2}}}{{\sin }^{-1}}x+x+C$.

Note: Whenever we get problems involving integration of logarithmic and inverse trigonometric functions, we make use of the uv rule of integration to reduce the calculation time. We should perform every step carefully in order to avoid calculation mistakes in the problem. We should not forget to add constant integration while solving problems which involve indefinite integrals as it is the most common mistake done by students.