I am stuck with the following question in my homework. I have already done a part of this question to get the reduction formula of this integral. $$\int^{\pi/2}_0 \cos^{2n}x\,dx$$ Which should be (I hope this is correct) $$I_n=\frac{(2n-1)I_{n-2}}{2n}$$ How to get a reduction formula for this integral? The reduction formula I solved should be useful in some way. $$\int^{\pi/2}_0 x^2 \cos^{2n}x\,dx$$
Thanks so much.
The reduction formula for the first integral is $$I_n=\frac{2n-1}{2n}I_{n-1} $$ Notice that when the argument of $I$ is $n-1$, the exponent of the cosine is $2(n-1)=2n-2$. As for the second integral, denote $$\begin{align} G_n&=\int_0^{\pi/2}x^2\cos^{2n}x\,dx=\int_0^{\pi/2}x^2\cos^{2(n-1)}x\cos^2x\,dx \\ &=\int_0^{\pi/2}x^2\cos^{2(n-1)}x\left(1-\sin^2 x\right)\,dx \\ &=G_{n-1}-\int_0^{\pi/2}x^2\cos^{2(n-1)}x\sin^2 x\,dx\tag{1} \end{align}$$ Use integration by parts for the last integral: $$\begin{align} &\int_0^{\pi/2}x^2\cos^{2(n-1)}x\sin^2 x\,dx\\ & =-\frac{\cos^{2n-1}x}{2n-1}(x^2\sin x)\Big|_0^{\pi/2}+\frac{1}{2n-1}\int_0^{\pi/2}\cos^{2n-1}x\left(2x\sin x+x^2\cos x\right)\,dx \\ &= \frac{2}{2n-1}\int_0^{\pi/2}\cos^{2n-1}x\sin x\,dx+\frac{1}{2n-1}G_n \tag{2} \end{align}$$ And integration by parts again: $$\begin{align} \int_0^{\pi/2}\cos^{2n-1}x\sin x\,dx &= -\frac{\cos^{2n}x}{2n}x\Big|_0^{\pi/2}+\frac{1}{2n}\int_0^{\pi/2}\cos^{2n}x\,dx \\ &= \frac{1}{2n}I_n\tag{3} \end{align} $$ Substitute $(3)$ in $(2)$, then $(2)$ in $(1)$, then simplify, and we have (for $n\geq 1$) $$G_n=\frac{2n-1}{2n}G_{n-1}-\frac{1}{2n^2}I_n $$