Question

How do you integrate $\int{\sqrt{4-9{{x}^{2}}}}$ using trigonometric substitutions?

Answer 1

In the problem they have asked to use the trigonometric substitutions, by looking the given equation we will use the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. So, we will take the substitution $9{{x}^{2}}=4{{\sin }^{2}}y$. From this equation we will calculate the value of $x$ and $dx$ in terms of $y$ and $dy$, by using the algebraic formulas and differentiation formulas. Now we will convert the given equation in terms of $y$ and $dy$ by substituting the value of $x$ and $dx$ in the given equation. Now we will simplify the obtained equation by applying the trigonometric formula $\cos 2\theta =2{{\cos }^{2}}\theta -1$. After integrating the above value, we will get the result in terms of $y$, but we need to calculate the result in terms of $x$, so we will calculate the value of $y$ from our assumption i.e., $9{{x}^{2}}=4{{\sin }^{2}}y$ and substitute it in the obtained equation to get the result.

Complete step by step answer:
Given that, $\int{\sqrt{4-9{{x}^{2}}}}$.
Let us take the substitution $9{{x}^{2}}=4{{\sin }^{2}}y$.
From the above substitution the value of $x$ can be obtained by taking square root on both sides of the equation, then
$\begin{aligned} & \sqrt{9{{x}^{2}}}=\sqrt{4{{\sin }^{2}}y} \\\ & \Rightarrow 3x=2\sin y \\\ & \Rightarrow x=\dfrac{2}{3}\sin y....\left( \text{i} \right) \\\ \end{aligned}$
Differentiating the above value, then we will get
$dx=\dfrac{2}{3}\cos y.dy$
Now substituting the all values, we have in the given equation, then we will get
$\int{\sqrt{4-9{{x}^{2}}}}dx=\int{\sqrt{4-4{{\sin }^{2}}y}}\dfrac{2}{3}\cos ydy$
Taking $4$ common in the square root function, then we will get
$\Rightarrow \int{\sqrt{4-9{{x}^{2}}}}dx=\int{\dfrac{2}{3}\sqrt{4\left( 1-{{\sin }^{2}}y \right)}}\cos ydy$
From the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we have the value $1-{{\sin }^{2}}y={{\cos }^{2}}y$, then the above equation modified as
$\begin{aligned} & \Rightarrow \int{\sqrt{4-9{{x}^{2}}}dx}=\int{\dfrac{2}{3}\times 2\sqrt{{{\cos }^{2}}y}\cos y}dy \\\ & \Rightarrow \int{\sqrt{4-9{{x}^{2}}}dx}=\int{\dfrac{4}{3}{{\cos }^{2}}y}dy \\\ \end{aligned}$
From the trigonometric equation $\cos 2\theta =2{{\cos }^{2}}\theta -1$, then value of ${{\cos }^{2}}y=\dfrac{1+\cos 2y}{2}$, then the above equation is modified as
$\begin{aligned} & \Rightarrow \sqrt{4-9{{x}^{2}}}dx=\dfrac{4}{3}\int{\dfrac{1+\cos 2y}{2}dy} \\\ & \Rightarrow \sqrt{4-9{{x}^{2}}}dx=\dfrac{4}{3}\times \dfrac{1}{2}\left[ \int{1dy}+\int{\cos 2ydy} \right] \\\ \end{aligned}$
We have the integration formulas $\int{1.dy}=y+C$, $\int{\cos 2y}dy=\dfrac{1}{2}\sin 2y+C$, then we will get
$\begin{aligned} & \Rightarrow \sqrt{4-9{{x}^{2}}}dx=\dfrac{2}{3}\left[ y+\dfrac{1}{2}\sin 2y \right]+C \\\ & \Rightarrow \sqrt{4-9{{x}^{2}}}dx=\dfrac{2}{3}y+\dfrac{1}{3}\sin 2y+C \\\ \end{aligned}$
We have the trigonometric formula $\sin 2\theta =2\sin \theta .\cos \theta$, then we will have
$\Rightarrow \sqrt{4-9{{x}^{2}}}dx=\dfrac{2}{3}y+\dfrac{1}{3}\times 2\sin y\cos y+C$
From equation $\left( \text{i} \right)$, we have
$\begin{aligned} & x=\dfrac{2}{3}\sin y \\\ & \Rightarrow \sin y=\dfrac{3}{2}x...\left( \text{ii} \right) \\\ & \Rightarrow y={{\sin }^{-1}}\left( \dfrac{3}{2}x \right)...\left( \text{iii} \right) \\\ \end{aligned}$
From trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, the value of $\cos y$ will be
$\begin{aligned} & {{\cos }^{2}}y=1-{{\sin }^{2}}y \\\ & \Rightarrow \cos y=\sqrt{1-{{\left( \dfrac{3}{2}x \right)}^{2}}}....\left( \text{iv} \right) \\\ \end{aligned}$
From equations $\left( \text{ii} \right)$, $\left( \text{iii} \right)$, $\left( \text{iv} \right)$, the value of $\int{\sqrt{4-9{{x}^{2}}}}$ is
$\begin{aligned} & \Rightarrow \int{\sqrt{4-9{{x}^{2}}}dx}=\dfrac{2}{3}{{\sin }^{-1}}\left( \dfrac{3}{2}x \right)+\dfrac{2}{3}\left( \dfrac{3}{2}x \right)\left( \sqrt{1-\dfrac{9}{4}{{x}^{2}}} \right)+C \\\ & \Rightarrow \int{\sqrt{4-9{{x}^{2}}}dx}=\dfrac{2}{3}{{\sin }^{-1}}\left( \dfrac{3}{2}x \right)+\dfrac{1}{2}x\sqrt{4-9{{x}^{2}}}+C \\\ \end{aligned}$

Hence the value of $\int{\sqrt{4-9{{x}^{2}}}dx}$ is $\dfrac{x}{2}\sqrt{4-9{{x}^{2}}}+\dfrac{2}{3}{{\sin }^{-1}}\left( \dfrac{3}{2}x \right)+C$

Note: In the problem they have specially mentioned to use the trigonometric substitution, so we have followed the above procedure otherwise we have direct formula for this problem i.e., $\int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx}=\dfrac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\dfrac{{{a}^{2}}}{2}{{\sin }^{-1}}\left( \dfrac{x}{a} \right)+C$. We will convert the given equation in the above form and use this formula to get the result.