Integrate $\int\frac{x^4}{\sqrt{a^2-x^2}}dx$ using trigonometric substitution 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.
 A: There is a typo, though it was not critical:
$$sin2\theta=2sin\theta cos\theta=2*\frac{x}{a}*\frac{\sqrt{\color{red}x^2-\color{red}a^2}}{a} \text{ must be } \\
\sin2\theta=2\sin\theta \cos\theta=2*\frac{x}{a}*\frac{\sqrt{\color{red}a^2-\color{red}x^2}}{a}$$
You made some minor errors while substituting:
$$\frac{a^4}{8}[3sin^{-1}\frac{x}{a}-\frac{\color{red}8x\sqrt{a^2-x^2}}{a^2}+\frac{\color{red}4x(\sqrt{a^2-x^2}^3}{a^\color{red}2}-\frac{\color{red}4x^3\sqrt{a^2-x^2}}{a^\color{red}2}]+C \text{ must be}\\
\frac{a^4}{8}[3\sin^{-1}\frac{x}{a}-\frac{\color{red}4x\sqrt{a^2-x^2}}{a^2}+\frac{x(\sqrt{a^2-x^2}\color{red})^3}{a^\color{red}4}-\frac{x^3\sqrt{a^2-x^2}}{a^\color{red}4}]+C=\\
\frac{3a^4}{8}\sin^{-1}\frac{x}{a}-\frac{xa^2\sqrt{a^2-x^2}}{2}+\frac{x(a^2-x^2)\sqrt{a^2-x^2}}{8}-\frac{x^3\sqrt{a^2-x^2}}{8}+C=\\
\frac{3a^4}{8}\sin^{-1}\frac{x}{a}-\frac{3xa^2\sqrt{a^2-x^2}}{8}-\frac{x^3\sqrt{a^2-x^2}}{4}+C$$
A: You need to substitute $\theta=\arcsin(\frac{x}{a})$, or as you said just find $\sin(4\theta) $ and $\sin (2\theta)$ using the double angle properties and put the values in the final answer.
