The value of the following integral $\int\limits_{0}^{2}{{{x}^{3}}\sqrt{2x-{{x}^{2}}}dx}$ is equal to:
(a) $\dfrac{7\pi }{2}$
(b) $\dfrac{7\pi }{4}$
(c) $\dfrac{7\pi }{8}$
(d) $\dfrac{7\pi }{16}$
Answer
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Hint: First of all, make the perfect square of $2x-{{x}^{2}}$ by adding and subtracting 1 to it and make this expression as: $2x-{{x}^{2}}+1-1$. After perfect squaring, we will assume $1-x$ as $t$ and then differentiating both the sides to convert integration in x to “t”. While integrating, you will require following integration: $\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx}=\dfrac{x\sqrt{{{a}^{2}}-{{x}^{2}}}}{2}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\left( \dfrac{x}{a} \right)$.
Complete step by step answer:
The integral that we have to evaluate is given below:
$\int\limits_{0}^{2}{{{x}^{3}}\sqrt{2x-{{x}^{2}}}dx}$
In the above integral, we are going to make the expression written in square root as perfect square.
Making $2x-{{x}^{2}}$ as perfect square by adding and subtracting 1 we get,
$\begin{align}
& 2x-{{x}^{2}}+1-1 \\
& 1-\left( {{x}^{2}}+1-2x \right) \\
\end{align}$
We know that ${{x}^{2}}+1-2x$ is the identity expansion of ${{\left( 1-x \right)}^{2}}$ so using this relation in the above we get,
$1-{{\left( 1-x \right)}^{2}}$
Substituting $1-{{\left( 1-x \right)}^{2}}$ in place of $2x-{{x}^{2}}$ we get,
$\int\limits_{0}^{2}{{{x}^{3}}\sqrt{1-{{\left( 1-x \right)}^{2}}}dx}$
Let us assume $(1-x)$ as “t” we get,
$1-x=t$
Differentiating on both the sides we get,
$-dx=dt$
Now, substituting $-dx=dt$ and changing limits from 0 to 2 to 1 to -1 we get,
$\int\limits_{1}^{-1}{{{\left( 1-t \right)}^{3}}\sqrt{1-{{\left( t \right)}^{2}}}\left( -dt \right)}$
Now, removing the negative sign in the integral by interchanging the lower and upper limit we get,
$\int\limits_{-1}^{1}{{{\left( 1-t \right)}^{3}}\sqrt{1-{{t}^{2}}}dt}$
We know that the identity of ${{\left( 1-t \right)}^{3}}$ is equal to:
$\begin{align}
& {{\left( 1 \right)}^{3}}-3{{\left( 1 \right)}^{2}}\left( t \right)+3\left( 1 \right){{\left( t \right)}^{2}}-{{t}^{3}} \\
& =1-3t+3{{t}^{2}}-{{t}^{3}} \\
\end{align}$
So, substituting the above expansion in the above integral we get,
$\int\limits_{-1}^{1}{\left( 1-3t+3{{t}^{2}}-{{t}^{3}} \right)\sqrt{1-{{t}^{2}}}dt}$
Opening the bracket we get,
$\int\limits_{-1}^{1}{\sqrt{1-{{t}^{2}}}dt}-3\int\limits_{-1}^{1}{t\sqrt{1-{{t}^{2}}}dt}+3\int\limits_{-1}^{1}{{{t}^{2}}\sqrt{1-{{t}^{2}}}dt}-\int\limits_{-1}^{1}{{{t}^{3}}\sqrt{1-{{t}^{2}}}dt}$
Now, solving each of the above integral separately we get,
$\int\limits_{-1}^{1}{\sqrt{1-{{t}^{2}}}dt}$
We know that integration of $\sqrt{{{a}^{2}}-{{x}^{2}}}$ with respect to x we get,
$\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx}=\dfrac{x\sqrt{{{a}^{2}}-{{x}^{2}}}}{2}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\left( \dfrac{x}{a} \right)$
In the given integral, “a” is 1 so substituting “a” as 1 and “x” as “t” in the above we get,
\[\begin{align}
& \int{\sqrt{{{\left( 1 \right)}^{2}}-{{t}^{2}}}dt}=\dfrac{t\sqrt{{{\left( 1 \right)}^{2}}-{{t}^{2}}}}{2}+\dfrac{{{\left( 1 \right)}^{2}}}{2}{{\sin }^{-1}}\left( \dfrac{t}{1} \right) \\
& \Rightarrow \int{\sqrt{1-{{t}^{2}}}dt}=\dfrac{t\sqrt{1-{{t}^{2}}}}{2}+\dfrac{1}{2}{{\sin }^{-1}}\left( t \right) \\
\end{align}\]
Putting lower limit as -1 and upper limit as 1 in the above integral we get,
$\int\limits_{-1}^{1}{\sqrt{1-{{t}^{2}}}dt}=\left[ \dfrac{t\sqrt{1-{{t}^{2}}}}{2}+\dfrac{1}{2}{{\sin }^{-1}}\left( t \right) \right]_{-1}^{1}$
Applying lower and upper limit in the above equation we get,
\[\begin{align}
& \int\limits_{-1}^{1}{\sqrt{1-{{t}^{2}}}dt}=\left[ \left( \dfrac{\left( 1 \right)\sqrt{1-{{1}^{2}}}}{2}+\dfrac{1}{2}{{\sin }^{-1}}\left( 1 \right) \right)-\left( \dfrac{-1\sqrt{1-{{\left( -1 \right)}^{2}}}}{2}+\dfrac{1}{2}{{\sin }^{-1}}\left( -1 \right) \right) \right] \\
& \Rightarrow \int\limits_{-1}^{1}{\sqrt{1-{{t}^{2}}}dt}=\left[ \left( 0+\dfrac{1}{2}{{\sin }^{-1}}\left( 1 \right) \right)-\left( 0+\dfrac{1}{2}{{\sin }^{-1}}\left( -1 \right) \right) \right] \\
\end{align}\]
We know that, the value of:
$\begin{align}
& {{\sin }^{-1}}\left( 1 \right)=\dfrac{\pi }{2} \\
& {{\sin }^{-1}}\left( -1 \right)=-\dfrac{\pi }{2} \\
\end{align}$
Using the above relations in solving the above integral we get,
\[\begin{align}
& \int\limits_{-1}^{1}{\sqrt{1-{{t}^{2}}}dt}=\left[ \left( 0+\dfrac{1}{2}\left( \dfrac{\pi }{2} \right) \right)-\left( 0+\dfrac{1}{2}\left( -\dfrac{\pi }{2} \right) \right) \right] \\
& \Rightarrow \int\limits_{-1}^{1}{\sqrt{1-{{t}^{2}}}dt}=\left[ \dfrac{\pi }{4}+\dfrac{\pi }{4} \right] \\
& \Rightarrow \int\limits_{-1}^{1}{\sqrt{1-{{t}^{2}}}dt}=\dfrac{2\pi }{4}=\dfrac{\pi }{2} \\
\end{align}\]
Hence, we have solved the first integral as:
\[\int\limits_{-1}^{1}{\sqrt{1-{{t}^{2}}}dt}=\dfrac{\pi }{2}\]
Now, solving the second integral we get,
$3\int\limits_{-1}^{1}{t\sqrt{1-{{t}^{2}}}dt}$
We know that the integration of:
$\int{t\sqrt{1-{{t}^{2}}}dt}=-\dfrac{1}{3}{{\left( 1-{{t}^{2}} \right)}^{\dfrac{3}{2}}}+C$
Using the above indefinite integral in solving the second integral we get,
$3\left[ -\dfrac{1}{3}{{\left( 1-{{t}^{2}} \right)}^{\dfrac{3}{2}}}+C \right]_{-1}^{1}$
Applying lower and upper limit we get,
$\begin{align}
& 3\left[ \left( -\dfrac{1}{3}{{\left( 1-{{\left( -1 \right)}^{2}} \right)}^{\dfrac{3}{2}}} \right)-\left( -\dfrac{1}{3}{{\left( 1-{{\left( -1 \right)}^{2}} \right)}^{\dfrac{3}{2}}} \right) \right] \\
& =3\left[ 0-0 \right] \\
& =0 \\
\end{align}$
Hence, the evaluation of second integral is:
$3\int\limits_{-1}^{1}{t\sqrt{1-{{t}^{2}}}dt}=0$
Solving the third integral $3\int\limits_{-1}^{1}{{{t}^{2}}\sqrt{1-{{t}^{2}}}dt}$ we get,
The above integral is of the form of:
$\int{{{t}^{2}}\sqrt{1-{{t}^{2}}}dt}=\dfrac{1}{8}\left( {{\sin }^{-1}}t-t\sqrt{1-{{t}^{2}}}\left( 1-2{{t}^{2}} \right) \right)+C$
Applying lower limit as -1 and upper limit as 1 in the above integral we get,
\[\begin{align}
& \int\limits_{-1}^{1}{{{t}^{2}}\sqrt{1-{{t}^{2}}}dt}=\left[ \dfrac{1}{8}\left( {{\sin }^{-1}}t-t\sqrt{1-{{t}^{2}}}\left( 1-2{{t}^{2}} \right) \right)+C \right]_{-1}^{1} \\
& \Rightarrow \int\limits_{-1}^{1}{{{t}^{2}}\sqrt{1-{{t}^{2}}}dt}=\left[ \dfrac{1}{8}\left( {{\sin }^{-1}}1-1\sqrt{1-{{1}^{2}}}\left( 1-2{{\left( 1 \right)}^{2}} \right) \right)-\dfrac{1}{8}\left( {{\sin }^{-1}}\left( -1 \right)-\left( -1 \right)\sqrt{1-{{\left( -1 \right)}^{2}}}\left( 1-2{{\left( -1 \right)}^{2}} \right) \right) \right] \\
& \Rightarrow \int\limits_{-1}^{1}{{{t}^{2}}\sqrt{1-{{t}^{2}}}dt}=\left[ \dfrac{1}{8}\left( {{\sin }^{-1}}1-0 \right)-\dfrac{1}{8}\left( {{\sin }^{-1}}\left( -1 \right)-0 \right) \right] \\
& \Rightarrow \int\limits_{-1}^{1}{{{t}^{2}}\sqrt{1-{{t}^{2}}}dt}=\left[ \dfrac{1}{8}\left( \dfrac{\pi }{2} \right)-\dfrac{1}{8}\left( -\dfrac{\pi }{2} \right) \right] \\
& \Rightarrow \int\limits_{-1}^{1}{{{t}^{2}}\sqrt{1-{{t}^{2}}}dt}=\left[ \dfrac{1}{8}\left( \dfrac{\pi }{2} \right)+\dfrac{1}{8}\left( \dfrac{\pi }{2} \right) \right] \\
& \Rightarrow \int\limits_{-1}^{1}{{{t}^{2}}\sqrt{1-{{t}^{2}}}dt}=\dfrac{2\pi }{16}=\dfrac{\pi }{8} \\
\end{align}\]
Multiplying the above integral by 3 we get,
\[3\int\limits_{-1}^{1}{{{t}^{2}}\sqrt{1-{{t}^{2}}}dt}=\dfrac{3\pi }{8}\]
Hence, we have solved the third integral as follows:
\[3\int\limits_{-1}^{1}{{{t}^{2}}\sqrt{1-{{t}^{2}}}dt}=\dfrac{3\pi }{8}\]
Solving the fourth integral we get,
$\int\limits_{-1}^{1}{{{t}^{3}}\sqrt{1-{{t}^{2}}}dt}$
In the above integral, we are going to substitute ${{u}^{2}}$ in place of $1-{{t}^{2}}$ we get,
$1-{{t}^{2}}={{u}^{2}}$
Differentiating on both the sides we get,
$-2tdt=2udu$
2 will be cancelled on both the sides and we get,
$-tdt=udu$
Using the above relation, we are going to convert the fourth integral in terms of “u” we get,
Now, when we put the lower and upper limits of the fourth integral in the equation $1-{{t}^{2}}={{u}^{2}}$ we get,
Substituting -1 in place of “t” in the above equation we get,
$\begin{align}
& 1-{{\left( -1 \right)}^{2}}={{u}^{2}} \\
& \Rightarrow 0=u \\
\end{align}$
Then the lower limit in the new integral becomes 0. Now, to get the upper limit of the integral by substituting “t” as 1 in $1-{{t}^{2}}={{u}^{2}}$ we get,
$\begin{align}
& 1-{{\left( 1 \right)}^{2}}={{u}^{2}} \\
& \Rightarrow 0=u \\
\end{align}$
Hence, we have got the upper limit as 0. Now, the lower and upper limit becomes 0 so the evaluation of the fourth integral becomes 0.
Hence, the value of the integral $\int\limits_{-1}^{1}{{{t}^{3}}\sqrt{1-{{t}^{2}}}dt}$ is equal to 0.
Substituting the values of integrals that we have solved in the above which is equal to: $\int\limits_{-1}^{1}{\sqrt{1-{{t}^{2}}}dt}-3\int\limits_{-1}^{1}{t\sqrt{1-{{t}^{2}}}dt}+3\int\limits_{-1}^{1}{{{t}^{2}}\sqrt{1-{{t}^{2}}}dt}-\int\limits_{-1}^{1}{{{t}^{3}}\sqrt{1-{{t}^{2}}}dt}$ we get,
$\begin{align}
& \dfrac{\pi }{2}-0+\dfrac{3\pi }{8}-0 \\
& =\dfrac{4\pi +3\pi }{8} \\
& =\dfrac{7\pi }{8} \\
\end{align}$
Hence, the correct option is (c).
Note:
In the above problem, the place where you can make mistakes is the calculation mistake like in the options you can see only the denominator is varying while all the parts are the same. So, if you make any mistake in the denominator then it will cost a negative mark in the examination.
Another concept that we can take home is that if we have to adjust the negative sign in the integral then we can interchange the limits.
Complete step by step answer:
The integral that we have to evaluate is given below:
$\int\limits_{0}^{2}{{{x}^{3}}\sqrt{2x-{{x}^{2}}}dx}$
In the above integral, we are going to make the expression written in square root as perfect square.
Making $2x-{{x}^{2}}$ as perfect square by adding and subtracting 1 we get,
$\begin{align}
& 2x-{{x}^{2}}+1-1 \\
& 1-\left( {{x}^{2}}+1-2x \right) \\
\end{align}$
We know that ${{x}^{2}}+1-2x$ is the identity expansion of ${{\left( 1-x \right)}^{2}}$ so using this relation in the above we get,
$1-{{\left( 1-x \right)}^{2}}$
Substituting $1-{{\left( 1-x \right)}^{2}}$ in place of $2x-{{x}^{2}}$ we get,
$\int\limits_{0}^{2}{{{x}^{3}}\sqrt{1-{{\left( 1-x \right)}^{2}}}dx}$
Let us assume $(1-x)$ as “t” we get,
$1-x=t$
Differentiating on both the sides we get,
$-dx=dt$
Now, substituting $-dx=dt$ and changing limits from 0 to 2 to 1 to -1 we get,
$\int\limits_{1}^{-1}{{{\left( 1-t \right)}^{3}}\sqrt{1-{{\left( t \right)}^{2}}}\left( -dt \right)}$
Now, removing the negative sign in the integral by interchanging the lower and upper limit we get,
$\int\limits_{-1}^{1}{{{\left( 1-t \right)}^{3}}\sqrt{1-{{t}^{2}}}dt}$
We know that the identity of ${{\left( 1-t \right)}^{3}}$ is equal to:
$\begin{align}
& {{\left( 1 \right)}^{3}}-3{{\left( 1 \right)}^{2}}\left( t \right)+3\left( 1 \right){{\left( t \right)}^{2}}-{{t}^{3}} \\
& =1-3t+3{{t}^{2}}-{{t}^{3}} \\
\end{align}$
So, substituting the above expansion in the above integral we get,
$\int\limits_{-1}^{1}{\left( 1-3t+3{{t}^{2}}-{{t}^{3}} \right)\sqrt{1-{{t}^{2}}}dt}$
Opening the bracket we get,
$\int\limits_{-1}^{1}{\sqrt{1-{{t}^{2}}}dt}-3\int\limits_{-1}^{1}{t\sqrt{1-{{t}^{2}}}dt}+3\int\limits_{-1}^{1}{{{t}^{2}}\sqrt{1-{{t}^{2}}}dt}-\int\limits_{-1}^{1}{{{t}^{3}}\sqrt{1-{{t}^{2}}}dt}$
Now, solving each of the above integral separately we get,
$\int\limits_{-1}^{1}{\sqrt{1-{{t}^{2}}}dt}$
We know that integration of $\sqrt{{{a}^{2}}-{{x}^{2}}}$ with respect to x we get,
$\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx}=\dfrac{x\sqrt{{{a}^{2}}-{{x}^{2}}}}{2}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\left( \dfrac{x}{a} \right)$
In the given integral, “a” is 1 so substituting “a” as 1 and “x” as “t” in the above we get,
\[\begin{align}
& \int{\sqrt{{{\left( 1 \right)}^{2}}-{{t}^{2}}}dt}=\dfrac{t\sqrt{{{\left( 1 \right)}^{2}}-{{t}^{2}}}}{2}+\dfrac{{{\left( 1 \right)}^{2}}}{2}{{\sin }^{-1}}\left( \dfrac{t}{1} \right) \\
& \Rightarrow \int{\sqrt{1-{{t}^{2}}}dt}=\dfrac{t\sqrt{1-{{t}^{2}}}}{2}+\dfrac{1}{2}{{\sin }^{-1}}\left( t \right) \\
\end{align}\]
Putting lower limit as -1 and upper limit as 1 in the above integral we get,
$\int\limits_{-1}^{1}{\sqrt{1-{{t}^{2}}}dt}=\left[ \dfrac{t\sqrt{1-{{t}^{2}}}}{2}+\dfrac{1}{2}{{\sin }^{-1}}\left( t \right) \right]_{-1}^{1}$
Applying lower and upper limit in the above equation we get,
\[\begin{align}
& \int\limits_{-1}^{1}{\sqrt{1-{{t}^{2}}}dt}=\left[ \left( \dfrac{\left( 1 \right)\sqrt{1-{{1}^{2}}}}{2}+\dfrac{1}{2}{{\sin }^{-1}}\left( 1 \right) \right)-\left( \dfrac{-1\sqrt{1-{{\left( -1 \right)}^{2}}}}{2}+\dfrac{1}{2}{{\sin }^{-1}}\left( -1 \right) \right) \right] \\
& \Rightarrow \int\limits_{-1}^{1}{\sqrt{1-{{t}^{2}}}dt}=\left[ \left( 0+\dfrac{1}{2}{{\sin }^{-1}}\left( 1 \right) \right)-\left( 0+\dfrac{1}{2}{{\sin }^{-1}}\left( -1 \right) \right) \right] \\
\end{align}\]
We know that, the value of:
$\begin{align}
& {{\sin }^{-1}}\left( 1 \right)=\dfrac{\pi }{2} \\
& {{\sin }^{-1}}\left( -1 \right)=-\dfrac{\pi }{2} \\
\end{align}$
Using the above relations in solving the above integral we get,
\[\begin{align}
& \int\limits_{-1}^{1}{\sqrt{1-{{t}^{2}}}dt}=\left[ \left( 0+\dfrac{1}{2}\left( \dfrac{\pi }{2} \right) \right)-\left( 0+\dfrac{1}{2}\left( -\dfrac{\pi }{2} \right) \right) \right] \\
& \Rightarrow \int\limits_{-1}^{1}{\sqrt{1-{{t}^{2}}}dt}=\left[ \dfrac{\pi }{4}+\dfrac{\pi }{4} \right] \\
& \Rightarrow \int\limits_{-1}^{1}{\sqrt{1-{{t}^{2}}}dt}=\dfrac{2\pi }{4}=\dfrac{\pi }{2} \\
\end{align}\]
Hence, we have solved the first integral as:
\[\int\limits_{-1}^{1}{\sqrt{1-{{t}^{2}}}dt}=\dfrac{\pi }{2}\]
Now, solving the second integral we get,
$3\int\limits_{-1}^{1}{t\sqrt{1-{{t}^{2}}}dt}$
We know that the integration of:
$\int{t\sqrt{1-{{t}^{2}}}dt}=-\dfrac{1}{3}{{\left( 1-{{t}^{2}} \right)}^{\dfrac{3}{2}}}+C$
Using the above indefinite integral in solving the second integral we get,
$3\left[ -\dfrac{1}{3}{{\left( 1-{{t}^{2}} \right)}^{\dfrac{3}{2}}}+C \right]_{-1}^{1}$
Applying lower and upper limit we get,
$\begin{align}
& 3\left[ \left( -\dfrac{1}{3}{{\left( 1-{{\left( -1 \right)}^{2}} \right)}^{\dfrac{3}{2}}} \right)-\left( -\dfrac{1}{3}{{\left( 1-{{\left( -1 \right)}^{2}} \right)}^{\dfrac{3}{2}}} \right) \right] \\
& =3\left[ 0-0 \right] \\
& =0 \\
\end{align}$
Hence, the evaluation of second integral is:
$3\int\limits_{-1}^{1}{t\sqrt{1-{{t}^{2}}}dt}=0$
Solving the third integral $3\int\limits_{-1}^{1}{{{t}^{2}}\sqrt{1-{{t}^{2}}}dt}$ we get,
The above integral is of the form of:
$\int{{{t}^{2}}\sqrt{1-{{t}^{2}}}dt}=\dfrac{1}{8}\left( {{\sin }^{-1}}t-t\sqrt{1-{{t}^{2}}}\left( 1-2{{t}^{2}} \right) \right)+C$
Applying lower limit as -1 and upper limit as 1 in the above integral we get,
\[\begin{align}
& \int\limits_{-1}^{1}{{{t}^{2}}\sqrt{1-{{t}^{2}}}dt}=\left[ \dfrac{1}{8}\left( {{\sin }^{-1}}t-t\sqrt{1-{{t}^{2}}}\left( 1-2{{t}^{2}} \right) \right)+C \right]_{-1}^{1} \\
& \Rightarrow \int\limits_{-1}^{1}{{{t}^{2}}\sqrt{1-{{t}^{2}}}dt}=\left[ \dfrac{1}{8}\left( {{\sin }^{-1}}1-1\sqrt{1-{{1}^{2}}}\left( 1-2{{\left( 1 \right)}^{2}} \right) \right)-\dfrac{1}{8}\left( {{\sin }^{-1}}\left( -1 \right)-\left( -1 \right)\sqrt{1-{{\left( -1 \right)}^{2}}}\left( 1-2{{\left( -1 \right)}^{2}} \right) \right) \right] \\
& \Rightarrow \int\limits_{-1}^{1}{{{t}^{2}}\sqrt{1-{{t}^{2}}}dt}=\left[ \dfrac{1}{8}\left( {{\sin }^{-1}}1-0 \right)-\dfrac{1}{8}\left( {{\sin }^{-1}}\left( -1 \right)-0 \right) \right] \\
& \Rightarrow \int\limits_{-1}^{1}{{{t}^{2}}\sqrt{1-{{t}^{2}}}dt}=\left[ \dfrac{1}{8}\left( \dfrac{\pi }{2} \right)-\dfrac{1}{8}\left( -\dfrac{\pi }{2} \right) \right] \\
& \Rightarrow \int\limits_{-1}^{1}{{{t}^{2}}\sqrt{1-{{t}^{2}}}dt}=\left[ \dfrac{1}{8}\left( \dfrac{\pi }{2} \right)+\dfrac{1}{8}\left( \dfrac{\pi }{2} \right) \right] \\
& \Rightarrow \int\limits_{-1}^{1}{{{t}^{2}}\sqrt{1-{{t}^{2}}}dt}=\dfrac{2\pi }{16}=\dfrac{\pi }{8} \\
\end{align}\]
Multiplying the above integral by 3 we get,
\[3\int\limits_{-1}^{1}{{{t}^{2}}\sqrt{1-{{t}^{2}}}dt}=\dfrac{3\pi }{8}\]
Hence, we have solved the third integral as follows:
\[3\int\limits_{-1}^{1}{{{t}^{2}}\sqrt{1-{{t}^{2}}}dt}=\dfrac{3\pi }{8}\]
Solving the fourth integral we get,
$\int\limits_{-1}^{1}{{{t}^{3}}\sqrt{1-{{t}^{2}}}dt}$
In the above integral, we are going to substitute ${{u}^{2}}$ in place of $1-{{t}^{2}}$ we get,
$1-{{t}^{2}}={{u}^{2}}$
Differentiating on both the sides we get,
$-2tdt=2udu$
2 will be cancelled on both the sides and we get,
$-tdt=udu$
Using the above relation, we are going to convert the fourth integral in terms of “u” we get,
Now, when we put the lower and upper limits of the fourth integral in the equation $1-{{t}^{2}}={{u}^{2}}$ we get,
Substituting -1 in place of “t” in the above equation we get,
$\begin{align}
& 1-{{\left( -1 \right)}^{2}}={{u}^{2}} \\
& \Rightarrow 0=u \\
\end{align}$
Then the lower limit in the new integral becomes 0. Now, to get the upper limit of the integral by substituting “t” as 1 in $1-{{t}^{2}}={{u}^{2}}$ we get,
$\begin{align}
& 1-{{\left( 1 \right)}^{2}}={{u}^{2}} \\
& \Rightarrow 0=u \\
\end{align}$
Hence, we have got the upper limit as 0. Now, the lower and upper limit becomes 0 so the evaluation of the fourth integral becomes 0.
Hence, the value of the integral $\int\limits_{-1}^{1}{{{t}^{3}}\sqrt{1-{{t}^{2}}}dt}$ is equal to 0.
Substituting the values of integrals that we have solved in the above which is equal to: $\int\limits_{-1}^{1}{\sqrt{1-{{t}^{2}}}dt}-3\int\limits_{-1}^{1}{t\sqrt{1-{{t}^{2}}}dt}+3\int\limits_{-1}^{1}{{{t}^{2}}\sqrt{1-{{t}^{2}}}dt}-\int\limits_{-1}^{1}{{{t}^{3}}\sqrt{1-{{t}^{2}}}dt}$ we get,
$\begin{align}
& \dfrac{\pi }{2}-0+\dfrac{3\pi }{8}-0 \\
& =\dfrac{4\pi +3\pi }{8} \\
& =\dfrac{7\pi }{8} \\
\end{align}$
Hence, the correct option is (c).
Note:
In the above problem, the place where you can make mistakes is the calculation mistake like in the options you can see only the denominator is varying while all the parts are the same. So, if you make any mistake in the denominator then it will cost a negative mark in the examination.
Another concept that we can take home is that if we have to adjust the negative sign in the integral then we can interchange the limits.
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