Integrate $\int\frac{x^4}{\sqrt{a^2-x^2}} \, dx$ using trigonometric substitution

Question

Integrate $\int\frac{x^4}{\sqrt{a^2-x^2}} \, dx$ using trigonometric substitution

Ok, so it's been a really long time since I've done a problem like this but after doing a little bit o studying, this is how far I've gotten.

$$\int\frac{x^4}{\sqrt{a^2-x^2}}\,dx$$

Focusing on the denominator and using a triangle, I found that $\sin\theta=\frac{x}{a}$, therefore $x=a\sin\theta$ and $dx=a\cos\theta \, d\theta$

Plugging this into the given problem, I got $\int\frac{x^4}{\sqrt{a^2-x^2}}\, dx$ ${}= \int\frac{a^4 \sin^4\theta}{a\cos\theta}\cdot a\cos\theta \, d \theta $= $a^4\int (\sin^4 \theta ) \, d \theta$

Since $\sin^2\theta=\frac{1-\cos2\theta}{2}$, I can say

$$a^4\int (\sin^4 \theta \,d \theta =a^4\int\frac{1-2\cos2\theta+\cos^22\theta}{4}\,)d\theta$$ Pulling out the 4 from the denominator and substituting $\cos^22\theta=1-\sin^22\theta$, I get

$$\frac{a^4}{4} \int({1-2\cos2\theta+1-\sin^22\theta})\, d\theta=\frac{a^4}{4}\int({2-2\cos2\theta-\sin^22\theta}) \, d\theta$$

Substituting $\sin^22\theta for \frac{1-\cos4\theta}{2}$ gets me

$$\frac{a^4}{4}\int(2-2\cos2\theta-\frac{1-\cos4\theta}{2})d\theta=\frac{a^4}{8}\int(4-4\cos2\theta-1+\cos4\theta) \, d\theta=\frac{a^4}{8}\int(3-4 \cos2\theta + \cos4\theta)\, d\theta$$

$$=\frac{a^4}{8}[3\theta-\frac{4\sin2\theta}{2}+\frac{\sin4\theta}{4}]+C$$

From here, there are some substitutions that we must do.

We know that $sin\theta=\frac{x}{a}$ So, $\theta=sin^{-1}(\frac{x}{a})$

Also, $sin2\theta=2sin\theta cos\theta$=$2*\frac{x}{a}*\frac{\sqrt{x^2-a^2}}{a}$ and $sin4\theta=4cos^3\theta sin\theta-4sin^3\theta cos\theta=4*({\frac{\sqrt{a^2-x^2}}{a}})^3*\frac{x}{a}-4*(\frac{x}{a})^3*\frac{\sqrt{a^2-x^2}}{a}$

Substituting this gives,

$\frac{a^4}{8}[3sin^{-1}\frac{x}{a}-\frac{8x\sqrt{a^2-x^2}}{a^2}+\frac{4x(\sqrt{a^2-x^2}^3}{a^2}-\frac{4x^3\sqrt{a^2-x^2}}{a^2}]+C$

And then finally, $\frac{3a^4}{8}sin^{-1}\frac{x}{a}-\frac{a^2x\sqrt{a^2-x^2}}{2}+\frac{a^2x(\sqrt{a^2-x^2}^3}{2}-\frac{ax^3\sqrt{a^2-x^2}}{2}]+C$

Is this correct? Please critique, comment, and help clarify. I really need to understand this problem.

Answer 1

This is a binomial differential integral of the type:

$I=\int x^{m}(A+B x^{n})^{q/r}dx$.

Chebiceff showed that:"such an integral can be traced back to an integral of rational function only when

$\frac{m+1}{n}$ or $\frac{m+1}{n}+\frac{q}{r}$, are integers.

The second case is the one that interests us. Let it be

$a^{2}-x^{2}=t^{2} x^{2}$,

$x=\Big(\frac{a^{2}}{t^{2}+1}\Big)^{(1/2)}$,

$dx=\frac{-t a}{(t^{2}+1)^{3/2}}dt$,

$ a^{2}-x^{2}=\frac{a^{2}t^{2}}{t^{2}+1}$,

$I=a^{4}\int \frac{1}{(t^{2}+1)^{3}}dt$,

then

$I=\frac{3a^{4}asin(\frac{x}{a})}{8}-\frac{x\sqrt{a^{2}-x^{2}}(2x^{2}+3a^{2})}{8}$, (Maple).

Answer 2

Let $x=a\cos\theta$ then ${dx/d\theta} = -a\sin\theta$ and $\sqrt{a^2-x^2}=a(1-(x/a)^2)^{1/2}=a\sin\theta$, so the integral

$$ \int {x^4\over\sqrt{a^2-x^2}} dx = -a^4\int\cos^4\theta \,d\theta. $$ Repeated substitution of the formula $\cos\phi = 1/2 (\cos 2\phi+1)$ gives $$ \int\cos^4\theta\,d\theta = {1\over 8}\int(\cos4\theta+4\cos2\theta+3)\,d\theta. $$ This integral is simply $$ {1\over4}(\sin4\theta+8\sin 2\theta+ 12\theta) + C $$ Use $\sin 2 \phi=2\sin\phi\cos\phi$ and $\cos 2\phi=2\cos^2\phi-1$ to get $$ 2\sin\theta\cos^3\theta+3\sin\theta\cos\theta+3\theta. $$ The answer for the whole integral is then $$ I={-a^4\over 8}(2qp^3+3qp+3\cos^{-1}(x/a))+C $$ where I've defined $p=x/a$ and $q=(1-p^2)^{1/2}$ to simplify.

To check this, differentiate with respect to x. Note that $$ {dq\over dx}={-x\over a^2 q} $$ and $$ {d\over dx}\cos^{-1}(x/a) = -{1\over aq} $$ Differentiating $I$ by $x$ gives $$ {dI\over dx}={-a^4\over 8}{1\over q}\left(-{2xp^3\over a^2}+{6x^2\over a^3}q^2-3{x^2\over a^3}+{3\over a}q^2-{3\over a}\right) $$ Then expand out $p=x/a$ and $q^2=1-(x/a)^2$. You'll find it all cancels out leaving the original integrand. I haven't included all the steps here since the post would get too long, but if you get stuck then leave a comment.