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I've tried to integrate $$\int\frac{1}{\sqrt{9-x^2}}dx$$ through substitution. I changed the equation to $$\int\frac{1}{\sqrt{9-x^2}}dx\equiv\int 1(9-x^2)^\frac{-1}2$$ and took $u=9-x^2$ whereby $du=-2xdx$ ($\frac{-du}{2x}=dx$) but then I got stuck because where do I go on from there? The answer is $\arcsin(\frac{1}{3}x)+c$, but how?

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5 Answers 5

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This is a form that we learn to recognize as related to inverse trigonometric functions. When you have an expression in your integrand that looks like $x^2\pm a^2$ or, as in this case, $a^2-x^2$ (for $a=3$), and you can't just use $u$-substitution, the strategy is to try trig substitution.

In this case, it's actually easier than that, because if you know that $\frac{d}{dx} \sin^{-1}x=\frac{1}{\sqrt{1-x^2}}$, then you realize that this integral can be transformed into that by juggling constants and doing a $u$-substitution of $u=\frac{x}{3}$.

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Substitute $x=3\sin\theta$ .....

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Hint:

let $$x=3\sin u$$ so $$dx=3\cos u du$$

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  • $\begingroup$ But why would I let $u=3\sin(x)$? Can you please elaborate on that? $\endgroup$
    – Ski Mask
    Jan 28, 2018 at 18:09
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    $\begingroup$ @SkiMask Recall the following identity $$\sin^2 (u)+\cos^2 (u)=1$$ This gives the motivation for this substitution. $\endgroup$ Jan 28, 2018 at 18:13
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write $$\sqrt{9-x^2}=3\sqrt{1-\left(\frac{x}{3}\right)^2}$$ and Substitute $$x=3t$$

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It's just $\arcsin\frac{x}{3}+C$ because it's the table integral.

Also, $$\left(\arcsin\frac{x}{3}\right)'=\frac{1}{\sqrt{1-\frac{x^2}{9}}}\cdot\frac{1}{3}=\frac{1}{\sqrt{9-x^2}}.$$

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