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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.

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  • $\begingroup$ 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? $\endgroup$ – bjorn93 Jan 21 at 14:25
  • $\begingroup$ 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 $\endgroup$ – c07091054 Jan 21 at 14:35
  • $\begingroup$ i have already solved the reduction formula for the first integral, but i am stuck for the second one $\endgroup$ – c07091054 Jan 21 at 15:23
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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 $n\geq 1$) $$G_n=\frac{2n-1}{2n}G_{n-1}-\frac{1}{2n^2}I_n $$

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  • $\begingroup$ Really thanks a lot for your help! Have a nice day, highly appreciated $\endgroup$ – c07091054 Jan 21 at 16:19
  • $\begingroup$ @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 $I_n$. $\endgroup$ – bjorn93 Jan 21 at 16:35
  • $\begingroup$ i understand that by now, thanks $\endgroup$ – c07091054 Jan 21 at 16:37

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