Integral substitution to find constant $\lambda$

analysiscalculusdefinite integralsintegrationreal-analysis

Whilst using the substitution $x = 4\sin^2\theta$, then show that this integral:

$$ \int^{3}_{0}\sqrt{\frac{x}{4-x}}dx $$

Is equivalent to:

$$ \int^{\frac{\pi}{3}}_{0}\sin^2\theta d\theta $$

What I have tried:

$$ \int^{3}_{0}\sqrt{\frac{x}{4-x}}dx=\left(\frac{x}{4-x}\right)^{\frac{1}{2}}=\left((4-x)^{\frac{1}{2}}\right)^{-x}=\int(4-x)^{-\frac{x}{2}}dx $$

Then substituting in $x = 4\sin^2\theta$

$$ \int(4-4\sin^2\theta)^{-2\sin^2\theta}=\int \frac{2\sin^2\theta}{4-4\sin^2\theta}d\theta=\int\frac{1}{2}-\frac{\sin^2\theta}{-2\sin^2\theta}d\theta $$

Have I got the right idea?

Second approach:

  1. $\int \sqrt{\frac{sin^2\theta}{1-sin^2\theta}}\cdot8\sin\theta\cos\theta d\theta$

identity:
$1-sin^2\theta = cos^2\theta$

  1. $\int \sqrt{\frac{sin^2\theta}{cos^2\theta}}=\sqrt{tan^2\theta}=\tan\theta$

Then using the this identity:
$\tan x=\frac{sinx}{cosx}$

  1. $8\int\frac{\sin\theta\sin\theta\cos\theta}{\cos \theta}d\theta=8\int sin^2\theta d\theta$

However, how do I get the upper integral to go from $3$ to $\frac{\pi}{3}$ likeso:
$\int^{3}_{0} \implies\int^{\frac{\pi}{3}}_{0}$?

Best Answer

First of all calculate the differential: $\mathrm{d}x = 8\sin\theta\cos\theta\,\mathrm{d}\theta$. Your boundaries are right. The integral is $$\int\limits_{0}^{\pi/3}\sqrt{\frac{4\sin^2\theta}{4(1-\sin^2\theta)}}{8\sin\theta\cos\theta}\,{\mathrm{d}\theta} = {8}\int\limits_{0}^{\pi/3}{\sin^2\theta}\,{\mathrm{d}\theta} = 8{\sqrt3},$$ since $\cos\theta= \sqrt{1-\sin^2\theta}$.

Edit 1: In your second approach you asked how to get to the boundaries. You want to substitute $x=4\sin^2\theta$, which means that $\theta=\arcsin\sqrt{x/4}$. Putting in your earlier values for $x$ in this formula gives you your new boundaries.

Edit 2: Solve for $\theta$: $$\begin{align} x&=4\sin^2\theta\quad\mid :4 \\\\ \frac{x}{4}&=\sin^2\theta\quad\mid \sqrt{} \\\\ \sqrt{\frac{x}{4}}&=\sin\theta\quad\mid \sqrt{} \\\\ \sqrt{\frac{x}{4}}&=\sin\theta\quad\mid \arcsin{} \\\\ \arcsin\sqrt{\frac{x}{4}}&=\theta \end{align}$$ Now you can set $x=0$ and $x=3$. That should give you $\theta=0$ and $\theta=\pi/3$.

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