Question

"find the reduction formula for the integral" sin^n(18x)

Answer 1

Let


`I(n,a)=\displaystyle\int\sin^nax\,\mathrm dx`
For demonstration on how to tackle this sort of problem, I'll only work through the case where
`n` is odd. We can write


`\displaystyle\int\sin^nax\,\mathrm dx=\int\sin^(n-2)ax\sin^2ax\,\mathrm dx=\int\sin^(n-2)ax(1-\cos^2ax)\,\mathrm dx`

`\implies I(n,a)=I(n-2,a)-\displaystyle\int\sin^(n-2)ax\cos^2ax\,\mathrm dx`

For the remaining integral, we can integrate by parts, taking


`u=\sin^(n-3)ax\implies\mathrm du=a(n-3)\sin^(n-4)ax\cos ax\,\mathrm dx`
`\mathrm dv=\sin ax\cos^2ax\,\mathrm dx\implies v=-\frac1{3a}\cos^3ax`


`\implies\displaystyle\int\sin^(n-2)ax\cos^2ax\,\mathrm dx=-\frac1{3a}\sin^(n-3)ax\cos^3ax+(a(n-3))/(3a)\int\sin^(n-4)ax\cos^4ax\,\mathrm dx`

For this next integral, we rewrite the integrand


`\sin^(n-4)ax\cos^4ax=\sin^(n-4)ax(1-\sin^2ax)^2=\sin^(n-4)ax-2\sin^(n-2)ax+\sin^nax`

`\implies\displaystyle\int\sin^(n-4)ax\cos^4ax\,\mathrm dx=I(n-4,a)-2I(n-2,a)+I(n,a)`

So putting everything together, we found


`I(n,a)=I(n-2,a)-\displaystyle\int\sin^(n-2)ax\cos^2ax\,\mathrm dx`

`I(n,a)=I(n-2,a)-\left(-\frac1{3a}\sin^(n-3)ax\cos^3ax+\frac{n-3}3\displaystyle\int\sin^(n-4)ax\cos^4ax\,\mathrm dx\right)`

`I(n,a)=I(n-2,a)-\frac{n-3}3\bigg(I(n-4,a)-2I(n-2,a)+I(n,a)\bigg)+\frac1{3a}\sin^(n-3)ax\cos^3ax`

`\frac n3I(n,a)=\frac{2n-3}3I(n-2,a)-\frac{n-3}3I(n-4,a)+\frac1{3a}\sin^(n-3)ax\cos^3ax`


`\implies I(n,a)=\frac{2n-3}nI(n-2,a)-\frac{n-3}nI(n-4,a)+\frac1{na}\sin^(n-3)ax\cos^3ax`