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$$x=\int \sqrt{\frac{y}{2a-y}}dy$$

According to my textbook, it says that the substitution by $y=a(1-\cos\theta)$ will easily solve the intergral. Why does this work?

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This doesn't answer your question, but this is #24 here: – Eric Angle Oct 18 '12 at 5:18

Let me try to answer this. First thing I'd do is try to rewrite the integral as


From here, I'd attempt to eliminate the square root by letting




Using a trigonometric identity, this can also be rewritten as


As for how to use the substitution as your book has it, though, you'll need to multiply numerator and denominator by $1-\cos\theta$. Continuing from where DonAntonio left off

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

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Find $\frac{dy}{d\theta}$ and then do the necessary replacement into the integral

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I think the OP might be asking how did one come up with this particular substitution in the first place – Hawk Oct 18 '12 at 4:45

$$y=a(1-\cos\theta)\Longrightarrow dy=a\sin\theta\,d\theta\Longrightarrow$$


But I can't see an easy way to solve the above, which according to WA is a rather ugly expression which, assuming positivity of everybody, equals $\,x+\sin x+C\,$, which would make the integrand equal to $\,1+\cos x\,$ . I can't see it right away but I guess there must be some trigonometric identity somewhere there (I already got the equality but not in a nice way).

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Of course...hehe. I differentiated both sides and got the same up to a constant...I must be going nuts. – DonAntonio Oct 18 '12 at 5:07

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