Question

Please help with this calculus problem!

Answer 1

`(\mathrm dQ)/(\mathrm dt)=(a(1-5bt^2))/((1+bt^2)^4)`

`Q(t)=\displaystyle\int(a(1-5bt^2))/((1+bt^2)^4)\,\mathrm dt`

Let
`t=\frac1{\sqrt b}\tan u`, so that
`\mathrm dt=\frac1{\sqrt b}\sec^2u\,\mathrm du`. Then the integral becomes


`\displaystyle\frac a{\sqrt b}\int\frac{1-5b\left(\frac1{\sqrt b}\tan u\right)^2}{\left(1+b\left(\frac1{\sqrtb}\tan u\right)^2\right)^4}\sec^2u\,\mathrm du`

`=\displaystyle\frac a{\sqrt b}\int(1-5\tan^2u)/((1+\tan^2u)^4)\sec^2u\,\mathrm du`

`=\displaystyle\frac a{\sqrt b}\int(1-5\tan^2u)/(\sec^6u)\,\mathrm du`

`=\displaystyle\frac a{\sqrt b}\int(\cos^6u-5\sin^2u\cos^4u)\,\mathrm du`

`=\displaystyle\frac a{\sqrt b}\int(6\cos^2u-5)\cos^4u\,\mathrm du`

One way to proceed from here is to use the power reduction formula for cosine:


`\cos^(2n)x=\left(\frac{1+\cos2x}2\right)^n`

You'll end up with


`=\displaystyle\frac a{16\sqrt b}\int(5\cos2u+8\cos4u+3\cos6u)\,\mathrm du`

`=\displaystyle\frac a{16\sqrt b}\left(\frac52\sin2u+2\sin4u+\frac12\sin6u\right)+C`

`=\frac a{32\sqrt b}(5\sin2u+4\sin4u+\sin6u)+C`

`Q(t)=(at)/((1+bt^2)^3)+C`