Solve definite Integral with given substitution

Question

I've been given the following integral

$$\int_a^b f(x) \,dx,$$

where $f(x) = \sqrt{16 - x^2}$ and $[a, b] = [-4, 4]$

I've been given the instruction to solve the definite integral with the following substitution: $$ x = 4\sin u $$

Various attempts of mine have failed. Help is very much appreciated.

Answer 1

$$\int_{-4}^4 \sqrt{16 - x^2}\,dx$$

Putting $x = 4 \sin \theta \implies dx = 4 \cos \theta$. And note that $\theta = \sin^{-1}\left(\frac x4\right)$.

Then $x = -4\implies \sin\theta = -1 \implies \theta = -\pi/2$, and $x = 4 \implies \sin\theta = 1 \implies \theta = \pi/2$.

$$\begin{align}\int_{-4}^4 \sqrt{16 - x^2}\,dx & = \int_{-\pi/2}^{\pi/2} \sqrt{16 - 16\sin^2 \theta}\,4\cos \theta\,d\theta\\ \\ & = 16\int_{-\pi/2}^{\pi/2} \sqrt{1 - \sin^2 \theta}\,\cos\theta\,d\theta \\ \\ \end{align}$$

Now you can use the identity that $1 - \sin^2 \theta = \cos^2\theta$.