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.

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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 $2\sin \theta$?

  • $\begingroup$ Are you familiar with the identity $\sin^2(\theta)=\frac{1-cos(2\theta)}{2}$? $\endgroup$ – 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?

  • $\begingroup$ 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. $\endgroup$ – Jim Sep 13 '14 at 20:23
  • $\begingroup$ which sqrt are you struggling with at the moment? I have also updated the hint with the integral. $\endgroup$ – Chinny84 Sep 13 '14 at 20:26
  • $\begingroup$ I understand the double angle bit and the integral you added in, but what I don't understand is how they go from $\displaystyle \int \sqrt{\dfrac{4\sin ^2 \theta}{4-4 \sin ^2 \theta}}\times 8\sin \theta \cos \theta d=\displaystyle \int 8\sin^2 \theta \text{d}\theta$ $\endgroup$ – Jim Sep 13 '14 at 20:29
  • $\begingroup$ I have updated the post now. $\endgroup$ – Chinny84 Sep 13 '14 at 20:35
  • $\begingroup$ Understand it all now! Thanks so much :) $\endgroup$ – Jim Sep 13 '14 at 20:36

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.


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