Question

Integral of 5xsqrt1-x^4

Answer 1

$\displaystyle\int5x\sqrt{1-x^4}\,\mathrm dx$

Take $x=\sqrt y$, so that $y^2=x^4$ and $\dfrac{\mathrm dy}{2\sqrt y}=\mathrm dx$. Note that this assumes $x\ge0$ and so requires $y\ge0$.

$\displaystyle5\int\sqrt y\sqrt{1-y^2}\dfrac{\mathrm dy}{2\sqrt y}=\frac52\int\sqrt{1-y^2}\,\mathrm dy$

Now take $y=\sin z$, so that $z=\arcsin y$. Note that for this invertibility condition to hold, we require that $-\dfrac\pi2\le z\le\dfrac\pi2$, which means $-1\le y\le1$. But since we already fixed $y\ge0$ with the previous substitution, we thus have $0\le y\le1$, which in turn restricts us to $0\le z\le\dfrac\pi2$. So with this substitution, we have $\mathrm dy=\cos z\,\mathrm dz$, which gives

$\displaystyle\frac52\int\sqrt{1-\sin^2z}\cos z\,\mathrm dz$
$=\displaystyle\frac52\int\sqrt{\cos^2z}\cos z\,\mathrm dz$
$=\displaystyle\frac52\int|\cos z|\cos z\,\mathrm dz$

Now over the interval $0$, we have $\cos z>0$, which means $|\cos z|=\cos z$, and so the integral is equivalent to

$\displaystyle\frac52\int\cos^2z\,\mathrm dz=\frac54\int(1+\cos2z)\,\mathrm dz$
$=\dfrac54z+\dfrac58\sin2z+C$
$=\dfrac54z+\dfrac54\sin z\cos z+C$
$=\dfrac54\arcsin y+\dfrac54\sin(\arcsin y)\cos(\arcsin y)+C$
$=\dfrac54\arcsin y+\dfrac54y\sqrt{1-y^2}+C$
$=\dfrac54\arcsin(x^2)+\dfrac54x^2\sqrt{1-x^4}+C$