Evaluate the following integral: $\int{\dfrac{{{t}^{4}}dt}{\sqrt{1-{{t}^{2}}}}}$.
Answer
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Hint: This integral can be solved by substituting t as a trigonometric function. Substitute t = $\sin x$. Then, use the formulas of integration to solve this question.
Complete step-by-step answer:
Before proceeding with the question, we must know all the formulas that will be required to solve this question.
In integration, we have a formula $\int{\cos nx=\dfrac{\sin nx}{n}}$ . . . . . . . . . . . . (1)
In trigonometry, we have a formula $\cos 2x=1-2{{\sin }^{2}}x$. From this formula, we can write,
${{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}$ . . . . . . . . . . . . . . . (2)
In trigonometry, we have a formula $\cos 2x=2{{\cos }^{2}}x-1$. From this formula, we can write,
\[{{\cos }^{2}}x=\dfrac{\cos 2x+1}{2}\] . . . . . . . . . . . . (3)
Also, in trigonometry, we have a formula $1-{{\sin }^{2}}x={{\cos }^{2}}x$. . . . . . . . . . . (4)
In algebra, we have a formula ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$. . . . . . . . (5)
In the question, we are required to evaluate $\int{\dfrac{{{t}^{4}}dt}{\sqrt{1-{{t}^{2}}}}}$. Let us substitute t = $\sin x$. Since t = $\sin x$, dt = $\cos xdx$.
$\Rightarrow \int{\dfrac{{{\sin }^{4}}x\cos xdx}{\sqrt{1-{{\sin }^{2}}x}}}$
Using formula (4), we can write it as,
\[\begin{align}
& \int{\dfrac{{{\sin }^{4}}x\cos xdx}{\sqrt{{{\cos }^{2}}x}}} \\
& \Rightarrow \int{\dfrac{{{\sin }^{4}}x\cos xdx}{\cos x}} \\
& \Rightarrow \int{{{\sin }^{4}}xdx} \\
& \Rightarrow \int{{{\left( {{\sin }^{2}}x \right)}^{2}}dx} \\
\end{align}\]
Using formula (2), we can write it as,
\[\begin{align}
& \int{{{\left( \dfrac{1-\cos 2x}{2} \right)}^{2}}dx} \\
& \Rightarrow \dfrac{1}{4}\int{{{\left( 1-\cos 2x \right)}^{2}}dx} \\
\end{align}\]
Using formula (5), we can write it as,
\[\begin{align}
& \dfrac{1}{4}\int{\left( 1+{{\cos }^{2}}2x-2\cos 2x \right)dx} \\
& \Rightarrow \dfrac{1}{4}\left( \int{1dx}+\int{{{\cos }^{2}}2xdx-2\int{\cos 2xdx}} \right) \\
\end{align}\]
Using formula (3), we can write \[{{\cos }^{2}}2x=\dfrac{\cos 4x+1}{2}\].
\[\begin{align}
& \Rightarrow \dfrac{1}{4}\left( \int{1dx}+\int{\left( \dfrac{\cos 4x+1}{2} \right)dx-2\int{\cos 2xdx}} \right) \\
& \Rightarrow \dfrac{1}{4}\int{1dx}+\dfrac{1}{4}\int{\left( \dfrac{\cos 4x+1}{2} \right)dx-\dfrac{1}{4}.2\int{\cos 2xdx}} \\
& \Rightarrow \dfrac{1}{4}\int{1dx}+\dfrac{1}{8}\int{\left( \cos 4x+1 \right)dx-\dfrac{1}{2}\int{\cos 2xdx}} \\
& \Rightarrow \dfrac{1}{4}\int{1dx}+\dfrac{1}{8}\int{\cos 4xdx}+\dfrac{1}{8}\int{1dx}-\dfrac{1}{2}\int{\cos 2xdx} \\
\end{align}\]
From formula (1), we can write \[\int{\cos 4xdx}=\dfrac{\sin 4x}{4}\] and \[\int{\cos 2xdx}=\dfrac{\sin 2x}{2}\]. Also, $\int{1dx}=x$. Substituting these integrals in the above integral, we get,
\[\dfrac{1}{4}x+\dfrac{1}{8}\dfrac{\sin 4x}{4}+\dfrac{1}{8}x-\dfrac{1}{2}\dfrac{\sin 2x}{2}\]
\[\Rightarrow \dfrac{3x}{8}+\dfrac{\sin 4x}{32}-\dfrac{\sin 2x}{4}\] . . . . . . . . . . . . (6)
In trigonometry, we have a formula.
$\sin 2x=2\sin x\cos x$
Using formula (4) in the above equation, we get,
$\sin 2x=2\sin x\sqrt{1-{{\sin }^{2}}x}$ . . . . . . . . (7)
Also, we have a formula $\sin 4x=2\sin 2x\cos 2x$. Substituting sin2x from formula (7) and $\cos 2x=1-2{{\sin }^{2}}x$ from formula (2), we get,
$\sin 4x=2\left( 2\sin x\sqrt{1-{{\sin }^{2}}x} \right)\left( 1-2{{\sin }^{2}}x \right)$ . . . . . . . . (8)
Substituting equation (7) and equation (8) in equation (6), we get,
\[\dfrac{3x}{8}+\dfrac{2\left( 2\sin x\sqrt{1-{{\sin }^{2}}x} \right)\left( 1-2{{\sin }^{2}}x \right)}{32}-\dfrac{2\sin x\sqrt{1-{{\sin }^{2}}x}}{4}\]
\[\Rightarrow \dfrac{3x}{8}+\dfrac{\left( 2\sin x\sqrt{1-{{\sin }^{2}}x} \right)\left( 1-2{{\sin }^{2}}x \right)}{16}-\dfrac{2\sin x\sqrt{1-{{\sin }^{2}}x}}{4}\] . . . . . . . . . . . . . . . . (9)
Since we had substituted t = sinx, substituting sinx = t and x = ${{\sin }^{-1}}t$, we get,
\[\begin{align}
& \dfrac{3{{\sin }^{-1}}t}{8}+\dfrac{\left( 2t\sqrt{1-{{t}^{2}}} \right)\left( 1-2{{t}^{2}} \right)}{16}-\dfrac{2t\sqrt{1-{{t}^{2}}}}{4} \\
& \Rightarrow \dfrac{3{{\sin }^{-1}}t}{8}+\dfrac{\left( 2t\sqrt{1-{{t}^{2}}} \right)}{4}\left( \dfrac{\left( 1-2{{t}^{2}} \right)}{4}-1 \right) \\
& \Rightarrow \dfrac{3{{\sin }^{-1}}t}{8}+\dfrac{\left( 2t\sqrt{1-{{t}^{2}}} \right)}{4}\left( \dfrac{\left( 1-2{{t}^{2}} \right)-4}{4} \right) \\
& \Rightarrow \dfrac{3{{\sin }^{-1}}t}{8}+\dfrac{\left( 2t\sqrt{1-{{t}^{2}}} \right)}{4}\left( \dfrac{-2{{t}^{2}}-3}{4} \right) \\
& \Rightarrow \dfrac{3{{\sin }^{-1}}t}{8}-\dfrac{\left( 2t\sqrt{1-{{t}^{2}}} \right)\left( 2{{t}^{2}}+3 \right)}{16} \\
\end{align}\]
Note: There is a possibility that one may commit a mistake while evaluating the integral of cosx. There is a possibility that one may write the integral of cosx as -sinx instead of sinx which may lead us to an incorrect answer.
Complete step-by-step answer:
Before proceeding with the question, we must know all the formulas that will be required to solve this question.
In integration, we have a formula $\int{\cos nx=\dfrac{\sin nx}{n}}$ . . . . . . . . . . . . (1)
In trigonometry, we have a formula $\cos 2x=1-2{{\sin }^{2}}x$. From this formula, we can write,
${{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}$ . . . . . . . . . . . . . . . (2)
In trigonometry, we have a formula $\cos 2x=2{{\cos }^{2}}x-1$. From this formula, we can write,
\[{{\cos }^{2}}x=\dfrac{\cos 2x+1}{2}\] . . . . . . . . . . . . (3)
Also, in trigonometry, we have a formula $1-{{\sin }^{2}}x={{\cos }^{2}}x$. . . . . . . . . . . (4)
In algebra, we have a formula ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$. . . . . . . . (5)
In the question, we are required to evaluate $\int{\dfrac{{{t}^{4}}dt}{\sqrt{1-{{t}^{2}}}}}$. Let us substitute t = $\sin x$. Since t = $\sin x$, dt = $\cos xdx$.
$\Rightarrow \int{\dfrac{{{\sin }^{4}}x\cos xdx}{\sqrt{1-{{\sin }^{2}}x}}}$
Using formula (4), we can write it as,
\[\begin{align}
& \int{\dfrac{{{\sin }^{4}}x\cos xdx}{\sqrt{{{\cos }^{2}}x}}} \\
& \Rightarrow \int{\dfrac{{{\sin }^{4}}x\cos xdx}{\cos x}} \\
& \Rightarrow \int{{{\sin }^{4}}xdx} \\
& \Rightarrow \int{{{\left( {{\sin }^{2}}x \right)}^{2}}dx} \\
\end{align}\]
Using formula (2), we can write it as,
\[\begin{align}
& \int{{{\left( \dfrac{1-\cos 2x}{2} \right)}^{2}}dx} \\
& \Rightarrow \dfrac{1}{4}\int{{{\left( 1-\cos 2x \right)}^{2}}dx} \\
\end{align}\]
Using formula (5), we can write it as,
\[\begin{align}
& \dfrac{1}{4}\int{\left( 1+{{\cos }^{2}}2x-2\cos 2x \right)dx} \\
& \Rightarrow \dfrac{1}{4}\left( \int{1dx}+\int{{{\cos }^{2}}2xdx-2\int{\cos 2xdx}} \right) \\
\end{align}\]
Using formula (3), we can write \[{{\cos }^{2}}2x=\dfrac{\cos 4x+1}{2}\].
\[\begin{align}
& \Rightarrow \dfrac{1}{4}\left( \int{1dx}+\int{\left( \dfrac{\cos 4x+1}{2} \right)dx-2\int{\cos 2xdx}} \right) \\
& \Rightarrow \dfrac{1}{4}\int{1dx}+\dfrac{1}{4}\int{\left( \dfrac{\cos 4x+1}{2} \right)dx-\dfrac{1}{4}.2\int{\cos 2xdx}} \\
& \Rightarrow \dfrac{1}{4}\int{1dx}+\dfrac{1}{8}\int{\left( \cos 4x+1 \right)dx-\dfrac{1}{2}\int{\cos 2xdx}} \\
& \Rightarrow \dfrac{1}{4}\int{1dx}+\dfrac{1}{8}\int{\cos 4xdx}+\dfrac{1}{8}\int{1dx}-\dfrac{1}{2}\int{\cos 2xdx} \\
\end{align}\]
From formula (1), we can write \[\int{\cos 4xdx}=\dfrac{\sin 4x}{4}\] and \[\int{\cos 2xdx}=\dfrac{\sin 2x}{2}\]. Also, $\int{1dx}=x$. Substituting these integrals in the above integral, we get,
\[\dfrac{1}{4}x+\dfrac{1}{8}\dfrac{\sin 4x}{4}+\dfrac{1}{8}x-\dfrac{1}{2}\dfrac{\sin 2x}{2}\]
\[\Rightarrow \dfrac{3x}{8}+\dfrac{\sin 4x}{32}-\dfrac{\sin 2x}{4}\] . . . . . . . . . . . . (6)
In trigonometry, we have a formula.
$\sin 2x=2\sin x\cos x$
Using formula (4) in the above equation, we get,
$\sin 2x=2\sin x\sqrt{1-{{\sin }^{2}}x}$ . . . . . . . . (7)
Also, we have a formula $\sin 4x=2\sin 2x\cos 2x$. Substituting sin2x from formula (7) and $\cos 2x=1-2{{\sin }^{2}}x$ from formula (2), we get,
$\sin 4x=2\left( 2\sin x\sqrt{1-{{\sin }^{2}}x} \right)\left( 1-2{{\sin }^{2}}x \right)$ . . . . . . . . (8)
Substituting equation (7) and equation (8) in equation (6), we get,
\[\dfrac{3x}{8}+\dfrac{2\left( 2\sin x\sqrt{1-{{\sin }^{2}}x} \right)\left( 1-2{{\sin }^{2}}x \right)}{32}-\dfrac{2\sin x\sqrt{1-{{\sin }^{2}}x}}{4}\]
\[\Rightarrow \dfrac{3x}{8}+\dfrac{\left( 2\sin x\sqrt{1-{{\sin }^{2}}x} \right)\left( 1-2{{\sin }^{2}}x \right)}{16}-\dfrac{2\sin x\sqrt{1-{{\sin }^{2}}x}}{4}\] . . . . . . . . . . . . . . . . (9)
Since we had substituted t = sinx, substituting sinx = t and x = ${{\sin }^{-1}}t$, we get,
\[\begin{align}
& \dfrac{3{{\sin }^{-1}}t}{8}+\dfrac{\left( 2t\sqrt{1-{{t}^{2}}} \right)\left( 1-2{{t}^{2}} \right)}{16}-\dfrac{2t\sqrt{1-{{t}^{2}}}}{4} \\
& \Rightarrow \dfrac{3{{\sin }^{-1}}t}{8}+\dfrac{\left( 2t\sqrt{1-{{t}^{2}}} \right)}{4}\left( \dfrac{\left( 1-2{{t}^{2}} \right)}{4}-1 \right) \\
& \Rightarrow \dfrac{3{{\sin }^{-1}}t}{8}+\dfrac{\left( 2t\sqrt{1-{{t}^{2}}} \right)}{4}\left( \dfrac{\left( 1-2{{t}^{2}} \right)-4}{4} \right) \\
& \Rightarrow \dfrac{3{{\sin }^{-1}}t}{8}+\dfrac{\left( 2t\sqrt{1-{{t}^{2}}} \right)}{4}\left( \dfrac{-2{{t}^{2}}-3}{4} \right) \\
& \Rightarrow \dfrac{3{{\sin }^{-1}}t}{8}-\dfrac{\left( 2t\sqrt{1-{{t}^{2}}} \right)\left( 2{{t}^{2}}+3 \right)}{16} \\
\end{align}\]
Note: There is a possibility that one may commit a mistake while evaluating the integral of cosx. There is a possibility that one may write the integral of cosx as -sinx instead of sinx which may lead us to an incorrect answer.
Last updated date: 22nd Sep 2023
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