Answer
405.3k+ views
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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)