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Are the steps below correct? Can you tell me another way to solve it, just hint?

$ \int \sqrt{16-x^2} \,dx $

$ x = 4 \sin u \rightarrow u=\arcsin(\frac{x}{4}) \rightarrow dx = 4\cos u \,dx$

$ \int \sqrt{16-x^2} \,dx = 16 \int \cos^2 u \,du = 16 \int \frac{\cos 2u +1}{2} \,du $

$ v = 2u \Rightarrow dv = 2 du \Rightarrow du = \frac{1}{2}dv $

$ 16 \int \frac{\cos 2u +1}{2} \,du = 16 \int \frac{\cos v}{2} +\frac{1}{2} \,dv = 16(\frac{\sin v}{2} +\frac{v}{2}) = 8\sin v +8v = 8\sin 2u +16u $

$ \Downarrow $

$8\sin(2\arcsin(\frac{x}{4})) + 4\arcsin\frac{x}{4} $

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    $\begingroup$ Just a minor remark: if you write $x=4\sin(u)$, then $dx$ can be calculated directly if you differentiate both sides with respect to $u$. Then you have $dx/du = 4\cos(u) \implies dx = 4\cos(u)du$. $\endgroup$
    – Joe
    Dec 19, 2020 at 14:12
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    $\begingroup$ On the other hand, it is more formally correct to write $u=\arcsin(x/4)$ when defining $u$ for the first time, rather than having $x=\ldots$. Also, in general $x/4=\sin(u)$ does not imply $\arcsin(x/4)=u$, since $\arcsin(\sin(\theta))\neq\theta$ (unless $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$). These technicalities are usually ignored when making substitutions though. $\endgroup$
    – Joe
    Dec 19, 2020 at 14:17

2 Answers 2

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Let $I = \int\sqrt{16-x^2} \ dx$. Then applying integration by parts with $u = \sqrt{16-x^2}, du = -\frac{x}{\sqrt{16-x^2}}$ and $dv = 1, v = x$:

$$I = x \sqrt{16-x^2} + \int \frac{x}{\sqrt{16-x^2}} \ dx$$

and now let $w = 16-x^2$.

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    $\begingroup$ How to go from Phoenix to New York passing from the Pacific first and the Atlantic then, but of course it was the question of the OP :) Nice way. $\endgroup$
    – Crash Bandicoot
    Dec 19, 2020 at 11:07
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The procedure is fine. Just be careful with the substitution in line 4, it is not necessary, but it is not wrong to do it. Nevertheless, you make a mistake since you forget to divide by 2.

Another way to solve it may be to substitute at the beginning for $x=4cos(u)$, the procedure is the same but formally is another method.

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