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Integral of 5xsqrt1-x^4

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\displaystyle\int5x√(1-x^4)\,\mathrm dx

Take
x=\sqrt y, so that
y^2=x^4 and
(\mathrm dy)/(2\sqrt y)=\mathrm dx. Note that this assumes
x\ge0 and so requires
y\ge0.


\displaystyle5\int\sqrt y√(1-y^2)(\mathrm dy)/(2\sqrt y)=\frac52\int√(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
-\frac\pi2\le z\le\frac\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\frac\pi2. So with this substitution, we have
\mathrm dy=\cos z\,\mathrm dz, which gives


\displaystyle\frac52\int√(1-\sin^2z)\cos z\,\mathrm dz

=\displaystyle\frac52\int√(\cos^2z)\cos z\,\mathrm dz

=\displaystyle\frac52\int|\cos z|\cos z\,\mathrm dz

Now over the interval
0<z<\frac\pi2, 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

=\frac54z+\frac58\sin2z+C

=\frac54z+\frac54\sin z\cos z+C

=\frac54\arcsin y+\frac54\sin(\arcsin y)\cos(\arcsin y)+C

=\frac54\arcsin y+\frac54y√(1-y^2)+C

=\frac54\arcsin(x^2)+\frac54x^2√(1-x^4)+C
User Jone
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