1. ## trig substitution integral

integral of sqrt(4-x^2)dx

I've gathered that x = 4sin(theta) so I write it out as:

sqrt(4 - 2sin(theta))2cos(theta)

that's supposed to equal 4cos^2(theta) but I am getting stuck in the middle here

2. Let $4\sin z=x \Rightarrow \cos z dz = dx$ so $\int\sqrt{4-x^2}dx=\int2\sqrt{1-\sin^2 z}\cos zdz=\int2\cos^2 zdz$ $=\int1+\cos 2zdz=z+\frac{\sin 2z}{2}+C=\arcsin \frac{x}{4}+\frac{x}{4}\left(1-\frac{x^2}{16}\right)+C$

3. Originally Posted by TYTY
integral of sqrt(4-x^2)dx

I've gathered that x = 4sin(theta) so I write it out as:

sqrt(4 - 2sin(theta))2cos(theta)

that's supposed to equal 4cos^2(theta) but I am getting stuck in the middle here
Letting $x = 2\sin\theta,$ for $-\frac\pi2\leq\theta\leq\frac\pi2,$ we have $dx=2\cos\theta\,d\theta$ and

$\int\sqrt{4-x^2}\,dx=\int2\cos\theta\sqrt{4-4\sin^2\theta}\,d\theta$

$=4\int\cos\theta\sqrt{1-\sin^2\theta}\,d\theta$

$=4\int\cos\theta\lvert\cos\theta\rvert\,d\theta$

$=4\int\cos^2\theta\,d\theta\qquad\bigg($since $\cos x\geq0\;\forall x\in\left[-\frac\pi2,\frac\pi2\right]\bigg)$

$=4\int\frac{1+\cos2\theta}2\,d\theta$

$=2\theta+\sin2\theta+C$

$=2\left(\arcsin\frac x2\right)+2\sin\theta\cos\theta+C$

$=2\arcsin\frac x2+\frac12x\sqrt{4-x^2}+C$

The reason we restrict $\theta$ to the interval $\left[-\frac\pi2,\frac\pi2\right]$ is so that there is a one-to-one correspondence between the values of $x$ and the values of $2\sin\theta.$

Also, be sure to notice that Abu-Khalil's solution is not quite correct: if you substitute $x=4\sin\theta,$ then the radicand becomes $4-16\sin^2\theta.$ However, I'm sure he only made the error because he was carrying over the faulty substitution from your original post.

4. Originally Posted by Abu-Khalil
Let $4\sin z=x \Rightarrow \cos z dz = dx$ so $\int\sqrt{4-x^2}dx=\int2\sqrt{1-\sin^2 z}\cos zdz=\int2\cos^2 zdz$ $=\int1+\cos 2zdz=z+\frac{\sin 2z}{2}+C=\arcsin \frac{x}{4}+\frac{x}{4}\left(1-\frac{x^2}{16}\right)+C$
Oops, I forgot $4$ when replacing $4\cos zdz$ and the square root at the end.