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

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

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

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

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

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

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

$$\Downarrow$$

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

• 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 $dx/du = 4\cos(u) \implies dx = 4\cos(u)du$.
– Joe
Dec 19, 2020 at 14:12
• 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.
– Joe
Dec 19, 2020 at 14:17

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 $$w = 16-x^2$$.
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.