# Prove the reduction formula for $\int\limits_{0}^{\pi/2} \sin^n x\, dx=\int\limits_{0}^{\pi/2} \cos^n x\, dx$

Problem: Prove that $$\int\limits_{0}^{\pi/2} \sin^n x\, dx=\int\limits_{0}^{\pi/2} \cos^n x\, dx=\begin{cases} \frac{n-1}{n}.\frac{n-3}{n-2}.\frac{n-5}{n-4}...\frac{2}{3}, & \mbox{ when n is odd } \\ \frac{n-1}{n}.\frac{n-3}{n-2}.\frac{n-5}{n-4}...\frac{1}{2}\frac{\pi}{2}, & \mbox{ when n is even}. \end{cases}$$

Let $$I_n = \int \sin^n{x} \ dx$$ \begin{align*} I_n &= -\cos x \hspace{3pt} \sin^{n-1}x + \int \cos^2 x \hspace{4pt} (n-1) \sin^{n-2} x \hspace{4pt} dx \\ &= -\cos x \hspace{3pt} \sin^{n-1}x + (n-1) \int \sin^{n-2} x \hspace{4pt} dx - (n-1) \int \sin^{n} dx\\ &= -\cos x \hspace{3pt} \sin^{n-1}x + (n-1) \int \sin^{n-2} x \hspace{4pt} dx - (n-1) I_n \end{align*}

$$\Rightarrow (1+n-1)I_n = -\cos x \hspace{3pt} \sin^{n-1}x + (n-1) \int \sin^{n-2} x dx$$

$$\Rightarrow I_n = \frac{-\cos x \hspace{3pt} \sin^{n-1}x}{n} + \frac{(n-1)}{n}I_{n-2}$$

• Integration by parts? Commented Dec 15, 2019 at 13:32
• What is your question ?
– user65203
Commented Dec 15, 2019 at 13:47

Using reduction formula for $$\sin(x)$$ we have:

$$\int_{0}^{\frac{\pi}{2}}\sin^{n}\left(x\right)dx=\frac{n-1}{n}\int_{0}^{\frac{\pi}{2}}\sin^{\left(n-2\right)}\left(x\right)dx-\frac{\sin^{\left(n-1\right)}\left(x\right)\cos\left(x\right)}{n} \Bigg|_0^\frac{\large\pi}{2}$$

Now a simple comupation shows that:

$$\frac{\sin^{\left(k-1\right)}\left(x\right)\cos\left(x\right)}{k}\Bigg|_0^\frac{\large\pi}{2}=\left(\frac{\sin^{\left(k-1\right)}\left(\frac{\pi}{2}\right)\cos\left(\frac{\pi}{2}\right)}{k}\right)-\left(\frac{\sin^{\left(k-1\right)}\left(0\right)\cos\left(0\right)}{k}\right)=0$$

for $$k\ge2$$

Finally the least even power which the reduction is valid for is $$k=2$$ ,setting $$k=2↦n$$ and substituting for $$n-1$$,$$n-2$$,...,$$2$$ respectively we get the same result, so I will prevent to write it again and again.

Continuing this way we have:$$\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdot\frac{n-5}{n-4}...\left[\int_{0}^{\frac{\pi}{2}}\sin^{\left(2\right)}\left(x\right)dx\right]$$$$=\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdot\frac{n-5}{n-4}...\left[\frac{2-1}{2}\int_{0}^{\frac{\pi}{2}}\sin^{0}\left(x\right)dx\right]$$$$=\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdot\frac{n-5}{n-4}...\frac{2-1}{2}\left[x\right]\Bigg|_0^\frac{\large\pi}{2}=$$$$\boxed {\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdot\frac{n-5}{n-4}\cdot...\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2}}$$ or $$\frac{\pi}{2}\prod_{k=1}^{\frac{n}{2}}\frac{2k-1}{2k}$$

So all we have to do is just a simple substitution. for the case $$n$$ odd just let $$k=3↦n$$ , (clearly the least odd power is $$3$$) ,then we have:

$$\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdot\frac{n-5}{n- 4}\cdot\cdot\cdot\left[\int_{0}^{\frac{\pi}{2}}\sin^{5}\left(x\right)dx\right]$$$$=\frac{n- 1}{n}\cdot\frac{n-3}{n-2}\cdot\frac{n-5}{n-4}\cdot\cdot\cdot\left[\frac{5- 1}{5}\int_{0}^{\frac{\pi}{2}}\sin^{3}\left(x\right)dx\right]$$$$=\frac{n-1}{n}\cdot\frac{n-3}{n- 2}\cdot\frac{n-5}{n-4}\cdot\cdot\cdot\frac{5-1}{5}\cdot\left[\frac{3- 1}{3}\int_{0}^{\frac{\pi}{2}}\sin^{1}\left(x\right)dx\right]=$$$$\boxed{\frac{n-1}{n}\cdot\frac{n-3}{n- 2}\cdot\frac{n-5}{n-4}\cdot...\cdot\frac{4}{5}\cdot\frac{2}{3}\cdot1}$$ or $$\prod_{k=1}^{\frac{n-1}{2}}\frac{2k}{2k+1}$$

and the result follows.

$$\dfrac{d(\sin^mx\cos x)}{dx}=-\sin^{m+1}x+m\sin^{m-1}x(1-\cos^2x)=m\sin^{m-1}x-(m+1)\sin^{m+1}x$$

Integrate both sides wrt $$x$$ between $$0,\dfrac\pi2$$

and write $$I_n=\int_0^{\pi/2}\sin^nx\ dx$$

$$mI_{m-1}=(m+1)I_{m+1}\iff I_{m+1}=?$$

$$I_0=?,I_1=?$$