Reduction formula for $\int_0^{\pi/2}\cos^m\theta\sin^n\theta\,d\theta$

Question

I have done the first section, but am totally stuck on how to proceed on (b).

Answer 1

\begin{align}J(m,n) =\int_0^{\frac{\pi}{2}} cos^m\theta sin^n\theta d\theta \\ J(m,n) =\int_0^{\frac{\pi}{2}} cos^{(m-1)}\theta sin^n\theta cos\theta d\theta\\ \end{align}

Then solve by parts where $sin^n\theta cos\theta = \frac{du}{dx}$ and in that integral substitute $u' = sin\theta$

\begin{align} \int u'^n \frac{du'}{dx} dx = \int u'^n du' = \frac{sin^{n+1}\theta}{n+1} + C \end{align}

And so:

\begin{align} J(m,n) &= \left[\frac{cos^{m-1}\theta sin^{n+1}\theta}{n+1}\right]_0^\frac{\pi}{2} - \int_0^{\frac{\pi}{2}}\frac{(m-1)cos^{(m-2)}\theta (-sin\theta)sin^{(n+1)}}{n+1} d\theta \\ &= 0 + \frac{m-1}{n+1} \int_0^{\frac{\pi}{2}} cos^{m-2}\theta(1-cos^2\theta)sin^n\theta d\theta\\ &= \frac{m-1}{n+1}J(m-2,n)-\frac{m-1}{n+1}J(m,n)\\ &=\frac{m-1}{n+m}J(m-2,n) \end{align}

Hope that helps

Answer 2

Try both ways. Write $$\sin(x) = -d(\cos(x))$$ and $$\cos(x) = d(\sin(x) )$$ Then integrate by parts!