# Find the value of $\int_0^1\sqrt{\frac{x}{4-x}}$ using $x=4\sin ^2 \theta$

Find the value of $\displaystyle \int_0^1\sqrt{\dfrac{x}{4-x}}$ using $x=4\sin ^2 \theta$

I'm trying to work through this, with the mark scheme, but I don't understand what they do.

I understand up to the line with the integral of $8\sin ^2\theta \ \text{d}\theta$:

Can someone explain to me the steps after that line? In particular, why are there two different integrals? I also don't really understand square root algebra when it comes to fractions - say I wanted to make the numerator 1 in the fraction, would I take out $2\sin \theta$?

• Are you familiar with the identity $\sin^2(\theta)=\frac{1-cos(2\theta)}{2}$? – graydad Sep 13 '14 at 20:30

$$\cos 2\theta =2\cos^2 \theta -1=1-2\sin^2\theta$$ \begin{align} \int 8\sin^2\theta d\theta &=& \int 8 \left(\dfrac{1-\cos 2\theta}{2}\right)d\theta \\ &=&\int4\left(1-\cos 2\theta\right)d\theta \end{align} the actually integral: $$\sqrt{\dfrac{4\sin^2\theta}{4-4\sin^2\theta}} = \sqrt{\dfrac{4}{4}}\sqrt{\dfrac{\sin^2\theta}{1-\sin^2\theta}} = \dfrac{\sin\theta}{\sqrt{\cos^2\theta}} = \dfrac{\sin \theta}{\cos \theta}$$
$$\int\sqrt{\dfrac{4\sin^2\theta}{4-4\sin^2\theta}} 8\cos \theta \sin \theta d\theta= \int \dfrac{\sin \theta}{\cos \theta} 8\cos \theta \sin \theta d\theta$$ can you take it from here?
• So the integral of $8\sin^2 \theta$ comes from them simplifying the expression? How exactly do you do it? I'm struggling with the algebra here, especially with that square root. – Jim Sep 13 '14 at 20:23
• I understand the double angle bit and the integral you added in, but what I don't understand is how they go from $\displaystyle \int \sqrt{\dfrac{4\sin ^2 \theta}{4-4 \sin ^2 \theta}}\times 8\sin \theta \cos \theta d=\displaystyle \int 8\sin^2 \theta \text{d}\theta$ – Jim Sep 13 '14 at 20:29
To answer your question on how they got from the first line to the second. $$\sqrt{ \dfrac{4\sin^2 \theta}{4-4\sin^2 \theta}} = \sqrt{ \dfrac{4}{4} \cdot \dfrac{\sin^2 \theta}{1-\sin^2 \theta} }$$ You should know a trigonometric identity to simplify the denominator in the argument of the sqrt.