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I have to integrate $$ \int \sqrt{x-x^2} dx $$ The answer on my textbook is $ \frac 14 \left( \arcsin(\sqrt x)-(1-2x)\sqrt{x-x^2} \right) $ but I want to solve this by myself so can you please give me a clue ? :) Thank you.

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did you try something already? does the solution has to be by parts (as the title suggests)? did you try some substitutions? If you are asking us for hints it will be better if you also tell us what you tried so far. You will get better hints. – Ittay Weiss Jan 26 '13 at 11:02

Hint: Complete the square: $$\int\sqrt{x-x^2}dx=\int\sqrt{-x^2+2\frac x2-\frac14+\frac14}dx=\int\sqrt{-\left(x-\frac 12\right)^2+\frac14}dx$$ Set $u=x-\frac 12$ to obtain $$\int\sqrt{-u^2+\frac14}du$$ I think you now how to deal with this one.

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For the real values of $\sqrt{x-x^2},$ we need $x-x^2\ge 0\implies 0\le x\le 1$

If we put $x=\sin^2t$ where $0\le t\le \frac\pi2, 1-x=\cos^2t\ge 0\implies \sqrt{1-x}=+\cos t$ and $dx=2\sin t\cos tdt$

So, $$\int\sqrt{x-x^2}dx=\int\sqrt{x(1-x)}dx$$ $$=\int\sin t\cos t2\sin t\cos tdt$$ $$=\frac12\int (\sin2t)^2dt \text{, as }\sin2y=2\sin y\cos y$$ $$=\frac14\int (1-\cos 4t)dt \text{, as } \cos2y=1-2\sin^2y$$ $$=\frac14\left(t-\frac{\sin4t}4\right)+c$$

Now, $x=\sin^2t\implies \cos2t=1-2\sin^2t=1-2x$ $\implies \sin2t=+\sqrt{1-(1-2x)^2}=2\sqrt{x-x^2}$ as $\sin2t\ge 0$ as $0\le 2t\le \pi$

So, $\sin4t=2\sin2t\cos2t=2\cdot 2\sqrt{x-x^2}(1-2x)$

So, $$\int\sqrt{x-x^2}dx=\frac14\left(\arcsin \sqrt x-4(1-2x)\sqrt{x-x^2}\right)+c$$

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You are such a genius.Mãĩ tumse pyār kartī hū̃. – gfdfd Jan 26 '13 at 11:09
If the down-vote is due to the fact that integration by parts is not used, the other answer did not follow it either. – lab bhattacharjee Jan 26 '13 at 11:12

for calculation of Integral $\displaystyle \int \sqrt{a^2-x^2}dx$

We will Use Integration by parts method

Let $\displaystyle \mathbb{I} = \int \sqrt{a^2-x^2}.xdx = \sqrt{a^2-x^2}.x-\int\frac{1}{2\sqrt{a^2-x^2}}.-2x.xdx$

$\displaystyle \mathbb{I}= x.\sqrt{a^2-x^2}-\int\frac{(a^2-x^2)-a^2}{\sqrt{a^2-x^2}}dx$

$\displaystyle \mathbb{I} = x.\sqrt{a^2-x^2}-\mathbb{I}+a^2.\sin^{-1}\left(\frac{x}{a}\right)$

So $\displaystyle \mathbb{I}=\sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}.\sin^{-1}\left(\frac{x}{a}\right)+\mathbb{C}$

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