Integral by trigonometric substitution: $ \int \:\frac{2y^3}{\sqrt{1-y^2}}\mathrm dy $

Question

$$ \int \:\frac{2y^3}{\sqrt{1-y^2}}\mathrm dy $$

I substitute $$ y = \sin t $$ then $$ 2 \int \:\frac{\sin^3 t \cdot \cos t}{\cos t}\;\mathrm dt = 2 \int \sin^2 t\;\mathrm d \cos t $$ What should I do next, and how do I convert back to the original variable $y$?

Answer 1

$$I=2\int\frac{x^3}{\sqrt{1-x^2}}\mathrm{d}x$$ $x=\sin u$: $$I=2\int\sin^3u\ \mathrm{d}u$$ $$I=2\int(1-\cos^2u)\sin u\ \mathrm{d}u$$ $w=\cos u$: $$I=2\int(w^2-1)\mathrm{d}w$$ $$I=2\int w^2\mathrm{d}w-2\int \mathrm{d}w$$ $$I=\frac23w^3-2w$$ $$I=\frac23\cos^3u-2\cos u$$ $$I=\frac23(1-x^2)^{3/2}-2(1-x^2)^{1/2}+C$$ $$I=-\frac23(2+x^2)\sqrt{1-x^2}+C$$

Answer 2

This type of problem is much easier if you do a $u-$sub by picking $u=1-y^2$ and $du=-2ydy,$ then

$$\int \frac{2y^3dy}{\sqrt{1-y^2}}=-\int\frac{(1-u)}{\sqrt{u}}du$$ and this becomes a simple power rule computation afterward.

Answer 3

You've established the following relationship.

$$\int{\frac{2y^3}{\sqrt{1-y^2}}dy}=\int{2\sin^2t\space dt}+C$$ Where $C$ is a fixed constant. You can evaluate the second integral using the rule $$\cos(2\alpha)=1-2\sin^2(\alpha)$$ Then use that $y=\sin t \to t=\sin^{-1}(y)$ to change the integral back.

Answer 4

Though I find the substitution $u=\sqrt{1-y^2}$ leading to the most elegant solution of this integral, you are asking for a help with the integral obtained through the substitution $y=\sin t.$

$$\begin{aligned}\int \:\frac{\sin^3 t \cdot \cos t}{\cos t}\;\mathrm dt& = \int \sin^2 t\;\mathrm d \cos t\\&=\int (1-\cos ^2 t)\;\mathrm d \cos t,\end{aligned}$$ which is easy to finish.
Then use $\:t=\arcsin y\:$ and $\;\cos (\arcsin y)=\sqrt{1-y^2}.$