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Having this:

$\int x\sqrt{1-x^2}dx$

Substitution:

$t = 1-x^2$

$dt = -2xdx => dx=\frac{-2x}{dt}$

So:

$$\int x\sqrt{1-x^2}dx = -\int x t^\frac{1}{2}\frac{2x}{dt} = -\int \frac{2x^2 t^\frac{1}{2}}{dt} = $$

...but here I stuck... I've tried $-4x\frac{t^\frac{-1}{2}}{\frac{-1}{2}} + c$ but it doesn't match correct result... why did I screw here?

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1  
$$dt = -2xdx => dx=\frac{-2x}{dt}$$ is wrong! –  Fermat May 12 at 18:02

3 Answers 3

up vote 1 down vote accepted

Use the fact that $\frac{d}{dx}((1-x^2)^\frac{3}{2})=-3x(1-x^2)^\frac{1}{2}$

(This is just using the rule $\frac{d}{dx}f(x)^n=f'(x)f(x)^{n-1}$)

And we have:

$\int x\sqrt{1-x^2}dx=(-1/3)\int 3x\sqrt{1-x^2}dx=(-1/3)\int \frac{d}{dx}((1-x^2)^\frac{3}{2})dx$

$=\int 1d((1-x^2)^\frac{3}{2})=-(1/3)(1-x^2)^\frac{3}{2}+c$

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Thanks ellya, though, could you be, please, more specific? I don't really understand the procedure you used. Thanks –  user984621 May 12 at 18:20
    
@user984621 I have put in an edit with a few more steps :) –  ellya May 12 at 18:25
    
You are not using here substitution? –  user984621 May 12 at 18:34
    
No, it's more using a bit of algebra, followed by the fundamental theorem of calculus, but it is similar to a substitution. –  ellya May 12 at 18:38

You've mistakenly flipped your fraction for $dx$. The fact that you have a $dt$ in the denominator should tip you off, as that symbol doesn't have meaning on its own.

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$$\int x\sqrt{1-x^2}dx=\dfrac{-1}{2}\int \sqrt{1-x^2}d(1-x^2)$$ Let $u=1-x^2$, and integrate. Note that if $-2xdx=dt$, then $xdx=\dfrac{-1}{2}dt$.

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