Definite integral of $y=\sqrt{(16-x^2)}$

Question

The integral is

$$\int^3_1\sqrt{16-x^2}dx$$

I've used the trig substitution method, replacing $x$ with $4\sin\theta$:

$$x=4\sin\theta, \quad \theta=\arcsin\left(\frac x4\right), \quad dx=4\cos\theta \ d\theta$$ (I've excluded the intervals for the definite integral for now) \begin{align} I&=\int\sqrt{16-16\sin^2\theta} \ 4\cos\theta d\theta \\ &=\int\sqrt{16\cos^2\theta}\ 4\cos\theta d\theta\\ &=\int4\cos\theta 4\cos\theta d\theta \\ &=16\int\cos^2\theta d\theta \\ &=16\int\frac{\cos2\theta}{2}d\theta \\ &=8\int \cos2\theta d\theta\\ &=\left[4\sin2\theta\right]^{\arcsin(3/4)}_{\arcsin(1/4)}\\ &=4\sin\left(2\arcsin\frac 34\right)-4\sin\left(2\arcsin\frac 14\right)\\ \end{align} At this last step, I'm not sure how to simplify the 2 arcsin part. It doesn't simplify because of the 2 in front of the arcsin, is there any other way? Or did I just do the wrong method to solve this integral in the first place?

Answer 1

First of all your double angle integral $$16\int(\frac{cos2\theta}{2})d\theta$$

should have been

$$ 16\int\frac {1}{2}(1+cos2\theta) d\theta$$

Towards the end of your solution, you asked for $$\sin(2 arcsin(1/4)).$$

Use double angle formula $$\sin(2x)=2 \sin(x) \cos(x)$$ and use the fact that $$\sin x =1/4.$$

Answer 2

Two things must be observed: $$\sqrt{\cos^2(t)}=|\cos(t)|$$ and $$\cos^2(t)=\frac{1}{2}\left(\cos(2t)+1\right)$$ Form your control, the result should be $$-1/2\,\sqrt {15}-8\,\arcsin \left( 1/4 \right) +3/2\,\sqrt {7}+8\, \arcsin \left( 3/4 \right) $$

Answer 3

$\sin(2\sin^{-1}(3/4))=2\sin(\sin^{-1}(3/4))\cos(\sin^{-1}(3/4))=2\cdot(3/4)\cdot(\sqrt{7}/4)$ because of the identity that $\cos(\sin^{-1}(3/4))=\sqrt{1-\sin^{2}(\sin^{-1}(3/4))}=\sqrt{1-9/16}=\sqrt{7}/4$.

Answer 4

Hint:

$$\sin(2\alpha)=2sin\alpha cos\alpha$$ where here $\alpha=\arcsin(1/4),\arcsin(1/4)$

and

$$\cos(\arcsin\beta)=\sqrt{1-\beta^2},\sin(\arcsin\beta)=\beta$$

where $\beta=1/4,3/4$