Courses
Courses for Kids
Free study material
Free LIVE classes
More

# Evaluate the following integral: $\int{\dfrac{{{t}^{4}}dt}{\sqrt{1-{{t}^{2}}}}}$.

Last updated date: 28th Mar 2023
Total views: 306.9k
Views today: 7.84k
Verified
306.9k+ views
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.

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.