# How to get a reduction formula for this integral?

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.

• So for which integral do you want a reduction formula? The first or the second? Also, is this a definite integral with limits of integration? – bjorn93 Jan 21 at 14:25
• Sorry for the vagueness, i was hoping to get the reduction formula for the second integral. The first and second integrals are actually both definite integral from 0 to pi/2. Correction is made – c07091054 Jan 21 at 14:35
• i have already solved the reduction formula for the first integral, but i am stuck for the second one – c07091054 Jan 21 at 15:23

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 $$n\geq 1$$) $$G_n=\frac{2n-1}{2n}G_{n-1}-\frac{1}{2n^2}I_n$$
• @c07091054 Because $I_{n-2}$ means that the exponent of the cosine is $2(n-2)=2n-4$. But we have to go from exponent $2n$ to exponent $2n-2$. Now if we had $$W_n=\int_0^{\pi/2}\cos^nx\,dx$$ then $$W_n=\frac{n-1}{n}W_{n-2}$$ It's basically a question of how we define $I_n$. – bjorn93 Jan 21 at 16:35