# Calculate $\int^1_{-1}\sqrt{(1-x^2)^3} dx$ using Integration by parts or substitution technique

How can I calculate $\int^1_{-1}\sqrt{(1-x^2)^3} dx$?.

I'm trying to solve it by using substitutions or integrate by parts but none of those leads me to the solution.

My attempt (with the help of a friend):

first since $(1-x^2)\ge 0$ than $\sqrt{(1-x^2)^3}$ could be written as $(1-x^2)^{2/3}$.

Now, $u=(1-x^2)^{2/3}, u' = -3x\sqrt{1-x^2}$. $v'=1, v =x$

and so $\int^1_{-1}\sqrt{(1-x^2)^3} dx = [x(1-x^2)^{3/2}]^1_{-1}+\int^1_{-1}3x^2\sqrt{1-x^2} dx$.

I don't even know if it's good, and if it is, I'm completely clueless at how to find $\int^1_{-1}3x^2\sqrt{1-x^2} dx$

• Put $x = \sin(t)$ ? Apr 3, 2017 at 17:39

"$1-x^2$" should immediately make you think about the substitution $x=\sin u$, as $1-x^2 = 1-\sin^2 u = \cos^2 u$. So let $x = \sin u$, then $dx = \cos u \,du$ and note that $\sin (-\tfrac{\pi}{2}) = -1$ and $\sin (\tfrac{\pi}{2}) = 1$, so \begin{align} \int_{-1}^{1} (1-x^2)^{3 / 2}\,dx &= \int_{-\pi/2}^{\pi/2} (1-\sin^2 u)^{3 / 2} \cos u\, du = \int_{-\pi/2}^{\pi/2} (\cos^2 u)^{3 / 2} \cos u\, du\\ &= \int_{-\pi/2}^{\pi/2} \cos^4 u\, du = \int_{-\pi/2}^{\pi/2} (\cos^2 u)^2\, du\\ &= \int_{-\pi/2}^{\pi/2} \left(\tfrac{1}{2} + \tfrac{1}{2} \cos(2u)\right)^2\, du\\ &= \int_{-\pi/2}^{\pi/2} \tfrac{1}{4} + \tfrac{1}{2}\cos(2u) + \tfrac{1}{4}\cos^2(2u)\; du\\ &= \int_{-\pi/2}^{\pi/2} \tfrac{1}{4} + \tfrac{1}{2}\cos(2u) + \tfrac{1}{4}\left(\tfrac{1}{2}+\tfrac{1}{2}\cos (4u) \right)\; du\\ &= \int_{-\pi/2}^{\pi/2} \tfrac{3}{8} + \tfrac{1}{2}\cos(2u) + \tfrac{1}{8}\cos (4u)\; du\\ \end{align} Now just use a bit of substitution to finish up.

Let $x=\sin(u)$ : $${\displaystyle\int}_{-1}^1x^2\sqrt{1-x^2}\,\mathrm{d}x={\displaystyle\int}_{-\pi/2}^{\pi/2}\sin^2(u)\cos(u)\sqrt{\cos^2(u)}\,\mathrm{d}u={\displaystyle\int}_{-\pi/2}^{\pi/2}\sin^2(u)\cos^2(u)\,\mathrm{d}u\\={\displaystyle\int}_{-\pi/2}^{\pi/2}\sin^2(2u)\,\mathrm{d}u$$

Let $x=2u$ : $$\class{steps-node}{\cssId{steps-node-1}{\dfrac{1}{2}}}{\displaystyle\int}\sin^2\left(x\right)\,\mathrm{d}x={\dfrac{1}{2}}{\displaystyle\int}\dfrac{1-\cos\left(2x\right)}{2}\,\mathrm{d}x$$

You should be able to finish.

Using the substitution $x = \sin(t), dx = \cos(t)dt$, we obtain the integral:

$$\int\limits_{-\pi/2}^{\pi/2} |cos(t)|^3 cos(t)dt = \int\limits_{-\pi/2}^{\pi/2} \cos^4(t)dt$$

Can you solve it from here?