How do I evaluate $\int \frac{x^4}{\sqrt{4-x^2}}dx$ using trig substitution?

Question

Here is what I have so far:

Let $x = 2\sin\theta$, and $dx=2\cos\theta \mathrm d\theta$

then the new integral is $$\int \frac{2^4\sin^4\theta}{\sqrt{4-4sin^2\theta}}2\cos\theta \mathrm d\theta$$

Then using the identity $1-\sin^2x = \cos^2x$ we can substitute $\cos^2\theta$ and simplify,

$$8\int \sin^4\theta d\theta$$

I then use the formula of $\int \sin^nudu= -\frac{1}{n}\sin^{n-1}u\cos + \frac{n-1}{n}\int \sin^{n-2}udu$, and I end up getting $$4(\theta-\sin^3\theta \cos\theta) +C$$

Is this correct?

How do I go about re-substituting $x$ into the equation?

Thanks in advance!

Answer 1

It should be$$ 16\int sin^4\theta d\theta $$ instead of $$8\int sin^4\theta d\theta$$ And after integration : $$\implies 6\theta - 4sin(2\theta) + \frac{1}{2}sin(4\theta) + C$$ Now, $$\theta = sin^{-1}\frac{x}{2}$$ Hence we can substitute this value of $$\theta$$ and get the final result

Answer 2

Also if $x=2\sin y$

WLOG $-\dfrac\pi2\le y\le\dfrac\pi2\implies\cos y\ge0$

$(2\sin^2y)^2=(1-\cos2y)^2=1-2\cos2y+\cos^22y$

and $2\cos^22y=1+\cos4y$

Now $\sin y=\dfrac x2,\cos y=+\dfrac{\sqrt{4-x^2}}2$ and $y=\arcsin\dfrac x2$