What's the integral of $f(x)=(1-x^2)^{1/2}$? I tried making $x=\sin(t)$ and doing integration by substitution but I don't think I arrived to the correct answer. All responses are appreciated...
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Perhaps it involves the use of arcsinx? – Vladimir Nabokov Oct 13 '13 at 12:21
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Putting $x=\sin\theta$
$$\int\sqrt{1-x^2}dx=\int\cos^2\theta d\theta=\int\frac12(1+\cos2\theta) d\theta$$
$$=\frac{\theta+\frac{\sin2\theta}2}2+C$$
$$=\frac{\theta+\sin\theta\cos\theta}2+C$$
$\theta=\arcsin x,\cos\theta=\sqrt{1-x^2}$
lab bhattacharjee
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Integrating by parts with $u(x)=\sqrt{1-x^2},dv(x)=1$
$$I=\int\sqrt{1-x^2}\cdot1dx=\sqrt{1-x^2}\int dx-\int\left(\frac{d(\sqrt{1-x^2})}{dx}\cdot \int dx\right)dx$$
$$=x\sqrt{1-x^2}-\int\frac{(-2x)}{2\sqrt{1-x^2}}xdx$$
$$=x\sqrt{1-x^2}+\int\frac{1-(1-x^2)}{\sqrt{1-x^2}} dx$$
$$=x\sqrt{1-x^2}+\int\frac1{\sqrt{1-x^2}} dx-I$$
$$\implies 2I=x\sqrt{1-x^2}+\arcsin x+C$$
lab bhattacharjee
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@JonathanY., have you noticed the $x$ with dx, which came as $\int dx$ – lab bhattacharjee Oct 13 '13 at 13:00
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Recall the three most common trigonometric substitutions and how they proceed.
Let $x=\sin\theta$. Then $\,dx = \cos\theta \,d\theta$. And $\theta = \arcsin x$.
$$\begin{align} \int\sqrt{1-x^2}\,dx & =\int\sqrt{1 - \sin^2\theta}\,\cos\theta \,d\theta \\ \\ & =\int\sqrt{\cos^2\theta}\,\cos\theta \,d\theta \\ \\ & = \int\cos^2\theta \,d\theta \\ \\ & = \int\frac12(1+\cos2\theta) \,d\theta\\ \\ & =\frac12\Big(\theta+\frac{\sin2\theta}2\Big)+C\\ \\ & =\frac12\Big(\theta+\sin\theta\cos\theta\Big)+C \\ \\ & = \frac 12 \Big(\arcsin x + x\sqrt{1-x^2}\Big) + C\tag{$\dagger$}\end{align}$$
$(\dagger)$. Since $x = \sin \theta$, $\theta = \arcsin x$. And $\cos x = \sqrt{1 - x^2}$.
How do we know $\cos x = \sqrt{1-x^2}\;?\;$
Note: $\sin^2 \theta + \cos ^2 \theta = 1 = x^2 + (\sqrt{1 - x^2})^2 = x^2 + 1 - x^2$.
amWhy
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