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I can solve this problem after $\int \sqrt{x}\sqrt{1-x} \text{dx}$ by substitution $x=u^2$ and then $u=\sin v$, and at last I get $\int \cos^4 x$.
But it looks a very long way, is there any shortest approach?

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3 Answers 3

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This may not be what you want, but the shortest way is definite by geometric intuition. The integrand is nothing but a semicircle centered at $(0.5, 0.0)$ with radius $0.5$.

So the integration, or the area, is $1/2 \cdot \pi (1/2)^2 = \pi/8$.semicircle

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  • $\begingroup$ It's beautiful! Of course after $x-x^2=1/4-(x-1/2)^2$ $\endgroup$
    – user217174
    May 5, 2016 at 7:16
  • $\begingroup$ @NNN, that's what math is! $\endgroup$
    – Taozi
    May 5, 2016 at 7:39
  • $\begingroup$ Yes, it's beautiful! $\endgroup$
    – user217174
    May 5, 2016 at 7:40
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$$\sqrt{x-x^2} = \frac{1}{2}\sqrt{1-(2x-1)^2},$$ so a single substitution of the form $$2x - 1 = \sin \theta, \quad dx = \frac{1}{2} \cos \theta \, d\theta$$ immediately yields $$\int \sqrt{x-x^2} \, dx = \frac{1}{4}\int \cos^2 \theta \, d\theta,$$ and the rest is straightforward.

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Hint:

By rewriting

$$\int \sqrt{x-x^2}dx=\int\sqrt{\frac14-\left(x-\frac12\right)^2}dx$$

Then use the substitution $$x-\frac12=\frac12\sin t$$

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  • $\begingroup$ Oh, yes! that's great. $\endgroup$
    – user217174
    May 5, 2016 at 7:08

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