Trig substitution for $\int \frac{x^2dx}{\sqrt{4 - x^2}}$

Question

According to my textbook the answer is: $2\arcsin(\frac{1}{2}x) - \sin(2\arcsin(\frac{1}{2}x))$, but I'm getting something slightly different.

First I tried setting $x = 2 \sin \theta, dx = 2 \cos \theta, \theta = \arcsin(\frac{x}{2})$

$$\int \frac{4\sin^2 \theta}{\sqrt{4(1-\sin^2 \theta})}*2\cos \theta d\theta $$

$$ \int \frac{4\sin^2 \theta}{2\cos \theta}2cos\theta = \int4\sin^2\theta{d\theta}$$

Then I let $u = 2\theta$ $$4 \int \frac{\sin 2\theta + 1}{2} = 4 [\frac{1}{2}\theta + \frac{1}{4} \int \sin udu]$$

$$4[\frac{1}{2}\theta - \frac{\cos u}{4}] = 4 [\frac{\arcsin(\frac{x}{2})}{2} - \cos(2\arcsin(\frac{x}{2}))]$$

Yielding: $$2\arcsin(\frac{x}{2}) - \frac{\cos(2\arcsin(\frac{x}{2}))}{4}$$

What's wrong with this?

Answer 1

You've apparently used an incorrect trigonometric identity,

$$\sin^2 t = \frac{1 + \sin 2t}{2}.$$

This is correct only if you erase all the sines and write cosines instead; it is false if $t = 0$, for example.

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The correct identity for $\sin^2 t$ is (among several others, of course)

$$\sin^2 t = \frac{1 - \cos 2t}{2}$$

leading to

$$\int \cos 2\theta \, d\theta = \frac 1 2 \sin 2\theta = \frac 1 2 \sin(2 \arcsin x/2)$$

which has the correct form.

Answer 2

I don't like your book's answer either.

$\sin(2\arcsin(\frac x{2}))$ should be simplified to $2\frac {x}{2}\sqrt {1-\frac {x^2}{4}} = \frac 12 x\sqrt {4-x^2}$

Others have pointed out the problem with the half-angle identity.