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Question

# Define $J(m, n)$, for non-negative integers $m$ and $n$, by the integral$J(m, n)=\int_0^{\pi / 2} \cos ^m \theta \sin ^n \theta d \theta .$(a) Evaluate $J(0,0), J(0,1), J(1,0), J(1,1), J(m, 1), J(1, n)$. (b) Using integration by parts, prove that, for $m$ and $n$ both $>1$,$J(m, n)=\frac{m-1}{m+n} J(m-2, n) \text { and } J(m, n)=\frac{n-1}{m+n} J(m, n-2) .$(c) Evaluate (i) $J(5,3)$, (ii) $J(6,5)$ and (iii) $J(4,8)$.

Solution

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$\textbf{Given}: J\left(m,n \right)=\int _{0}^{\frac{\pi}{2}} \cos^{m}\theta \sin^{n}\theta d\theta$

$\textbf{(a):} \text{Put al the values in above equation}$

\begin{align*} &J\left(0,0 \right)= \int _{0}^{\frac{\pi}{2}} \cos^{0}\theta \sin^{0}\theta d\theta \\ &=\int _{0}^{\frac{\pi}{2}} 1d\theta \\ &=\left[\theta \right]_{0}^{\frac{\pi}{2}}\\ &=\left(\dfrac{\pi}{2}-0 \right)\\ &=\dfrac{\pi}{2}\\ &J\left(0,1 \right)= \int _{0}^{\frac{\pi}{2}} \cos^{0}\theta \sin^{1}\theta d\theta \\ &=\int _{0}^{\frac{\pi}{2}} \sin\theta d\theta \\ &=\left[-\cos\theta \right]_{0}^{\frac{\pi}{2}}\\ &=-\left(\cos\dfrac{\pi}{2}-\cos0 \right)\\ &=-\left(0-1 \right)\\ &=1\\ &J\left(1,0 \right)= \int _{0}^{\frac{\pi}{2}} \cos^{1}\theta \sin^{0}\theta d\theta \\ &=\int _{0}^{\frac{\pi}{2}} \cos\theta d\theta \\ &=\left[\sin\theta \right]_{0}^{\frac{\pi}{2}}\\ &=\left(\sin\dfrac{\pi}{2}-\sin0 \right)\\ &=\left(1-0 \right)\\ &=1\\ \end{align*}

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