i used the substitution $$x=\frac{\sin{u}}{2}$$
and I got to $$\frac{1}{4}(\frac{1}{2}\sin{(2\arcsin(2x))}+\arcsin(2x))+c$$
and $$2x=\sin(u)$$
and drew a triangle
now im stuck...
the answer is
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lemonadedoge
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1Duplicate http://math.stackexchange.com/a/153851/243873 – VJP May 29 '15 at 04:15
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1use the fact that $\sin (2t) = 2 \sin t \cos t, \sin t = 2x, \cos t = \sqrt{-1-4x^2}$ to simplify $\sin(2 \sin^{-1}(2x))$ – abel May 29 '15 at 04:15
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1i meant to write $\cos t = \sqrt{1-4x^2}$ above. – abel May 29 '15 at 04:36
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$\frac{1}{2}\sin{(2\arcsin(2x))}=\frac{1}{2}\sin{2u}=\frac{1}{2}\sin{4u\sqrt{1-4u^2}}$ ...? can you answer it because i cant get it... @abel – lemonadedoge May 29 '15 at 04:52
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i posted some explanation. – abel May 29 '15 at 04:59
3 Answers
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You need to appeal to an identity at the end, before using the triangle:
$\int (1-4x^{2})^{1(/2)}dx=\int \cos ^{2}zdz=\int (\frac{1}{2}+\frac{1}{2}\cos 2z)dz=\frac{1}{2}z-\frac{1}{4}\sin 2z=\frac{1}{2}(z)-\frac{1}{4}(2\sin z\cos z)=\frac{1}{2}\arcsin x+\frac{1}{2}x(1-x^{2})^{1/2}+C$
Matematleta
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hint: let $u = \sqrt{1-4x^2}, v = x$, can you proceed with integration by parts ?
DeepSea
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You're lettng $v=x$ but there is no factor of $x$ in the integrand. Are we talking about the same integral? You could try lettng $v'=1$, and $u$ the root but I don't think that will work because in the second integral you get $x^{2}$ in the numerator and then the root in the denominator. Then you're back to trig substitution, aren't you? – Matematleta May 29 '15 at 04:35
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I'm getting $x(1-4x^{2})^{1/2}-4\int \frac{x^{2}}{(1-4x^{2})^{1/2}}dx$ by parts. So.... – Matematleta May 29 '15 at 04:41
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look at the first expression in you answer: you have $$\begin{align}\sin\left(2\sin^{-1}(2x)\right) &= \sin(2t)\\ & = 2\sin t \cos t \\ &=2 \times 2x\times \sqrt{1 -\sin^2 t}\\ &=4x\sqrt{1-4x^2}\end{align}$$ i have used $t = \sin^{-1}(2x), \sin t = 2x, \cos t = \sqrt{1-4x^2}.$
abel
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