Strange Trig Substitution in Integral

Question

Compute the following integral

$$\int _0^{0.6}\:\frac{x^2}{\sqrt{9-25x^2}}\mathrm dx$$

So as I have been doing in the past, I thought the required substitution would have been $x = 3\sin\theta$. But in this case, it is actually $x = \frac{3}{5}\sin\theta$. I am not entirely sure why this is. Is it because the coefficient of the $x^2$ in the denominator is also a square number so we can square root it and place it in the denominator of the substitution? Is this a universal rule?

Answer 1

In order to get rid of the square root in $$\int _0^{0.6}\:\frac{x^2}{\sqrt{9-25x^2}}dx$$

We have to have $9-25x^2$ as a perfect square. This is achieved by having $25x^2 = 9 \sin^2(\theta)$ which implies $9-25x^2 = 9 \cos^2(\theta)$, which is a perfect square.

Answer 2

Because $x=\frac{3}{5}\sin\theta$ implies $x^2=\frac{9}{25}\sin^2\theta$ and when you substitute that into the the integral, you get $9-9\sin^2\theta$ under the radical. Applying the pythagorean theorem results in $9\cos^2\theta$. (The suggested "move" is all about "facilitating" the pythagorean theorem essentially)

Answer 3

Maybe to make things clear for you, you need to perform two substitutions.

$$\begin{align} J=\int_0^{0.6}\frac{x^2}{\sqrt{9-25x^2}}\,dx &=\int_0^{0.6}\frac{x^2}{3\sqrt{1-\left(\frac{5}{3}x\right)^2}}\,dx \end{align}$$

Perform the change of variable $y=\frac{5}{3}x$,

$$\begin{align} J&=\frac{3}{5}\int_0^{1}\frac{\left(\frac{3}{5}y\right)^2}{3\sqrt{1-y^2}}\,dy\\ &=\frac{9}{125}\int_0^{1}\frac{y^2}{\sqrt{1-y^2}}\,dy \end{align}$$

Since $0.6\times \frac{5}{3}=1$.

The integral bounds belong to $[-1;1]$ and there is the factor $\dfrac{1}{\sqrt{1-y^2}}$ therefore you know the substitution $y=\sin u$ is a way to complete your computation.

Perform the substitution $y=\sin u$,

$$\begin{align} J&=\frac{9}{125}\int_0^{1}\frac{\sin^2 u}{\sqrt{1-\sin^2 u}}\times \cos u\,du\\ &=\frac{9}{125}\int_0^{1}\sin^2 u\,du\\ \end{align}$$

The substitution works because,

$$\begin{align} \frac{1}{\sqrt{1-y^2}}\,dy&=\frac{\cos u}{\sqrt{1-\sin^2 u}}\,du\\ &=\,du \end{align}$$

(the derivative of the sinus function is the cosinus function)

No more denominator.

To achieve the computation you need to linearize $\sin^2 u$.

If you know that

$$\begin{align}\cos(2u)&=\cos^2 u-\sin^ 2 u\\ &=\left(1-\sin^2 u\right) -\sin^2 u\\ &=1-2\sin^2 u\\ \end{align}$$

you are done.

But, there is an algorithmic way to tackle the question.

$$\begin{align}\sin^2 u&=\left(\frac{\text{e}^{iu}-\text{e}^{-iu}}{2i}\right)^2\\ &=-\frac{1}{4}\left(\text{e}^{i2u}+\text{e}^{-i2u}-2\times \text{e}^{iu}\times \text{e}^{-iu}\right)\\ &=-\frac{1}{4}\left(\text{e}^{i2u}+\text{e}^{-i2u}-2\right)\\ &=\frac{1}{2}-\frac{1}{2}\cos(2u) \end{align}$$

Since,

$$\displaystyle \sin u=\frac{\text{e}^{iu}-\text{e}^{-iu}}{2i},\cos u=\frac{\text{e}^{iu}+\text{e}^{-iu}}{2},i^2=-1$$

Therefore,

$$\begin{align}J&=\frac{9}{125}\int_0^1 \left(\frac{1}{2}-\frac{1}{2}\cos(2u)\right)\,du\\ &=\frac{9}{125}\Big[\frac{1}{2}u-\frac{\sin(2u)}{4}\Big]_{0}^{\frac{\pi}{2}}\\ &=\frac{9}{125}\times \frac{\pi}{4}\\ &=\boxed{\frac{9}{500}\pi} \end{align}$$

NB:

$\sqrt{1-x^2}$ does exist if and only if $x\in [-1;1]$

You can compute integral $\displaystyle \int_0^1 \frac{P(x)}{\sqrt{1-x^2}}\,dx$ , $P$ is a polynomial function.

Answer 4

Look at $\sqrt{9-25x^2}$ It would be nice to find a substitution that would make the square root go away. So we start by symplifying a bit

$$ \sqrt{9-25x^2} = 3 \sqrt{1 - \dfrac{25}{9}x^2} $$

Now compare $\sqrt{1 - \dfrac{25}{9}x^2}$ to $\sqrt{1 - \sin^2 \theta} = \cos{\theta}$

It seems that $x = \dfrac 35 \sin \theta$ might be a useful substitution.

Note that the biggest (most useful) domain about the origin for $\sin \theta$ is $\theta \in \left[ -\dfrac {\pi}{2}, \dfrac {\pi}{2} \right]$ since, then, $\cos \theta \in [0, 1]$. So, using $x = \dfrac 35 \cos\theta$ and getting $\sqrt{1 - \cos^2 \theta} = \sin{\theta}$ might cause problems.