$\displaystyle\int_0^\pi\sin^5\theta\cos^4\theta \,\mathrm d\theta=2\int_0^{\frac{\pi}{2}}\sin^5\theta\cos^4\theta \,\mathrm d\theta$

Question

My instructor wrote

$$\int_0^\pi\sin^5(\theta)\cos^4(\theta) \,\mathrm d\theta\tag{1}$$ $$=2\int_0^{\frac{\pi}{2}}\sin^5(\theta)\cos^4(\theta) \,\mathrm d\theta\tag{2}$$ $$=2\cdot\frac{\Gamma\left(\frac{5+1}{2}\right)\cdot\Gamma\left(\frac{4+1}{2}\right)}{2\cdot\Gamma\left(\frac{5+1}{2}+\frac{4+1}{2}\right)}\\=\frac{16}{315}.$$

How did he know that going from step $(1)$ to $(2)$ is valid? We can vaguely verify the step by checking the graph, which he had not seen.

Symmetry. Sine is the same on the left half of $[0,\pi]$ as is on the right half, and cosine is the opposite, but minus signs go away when you perform even powers. These facts show up over and over and over again, so your instructor was already familiar with this common trick. — coiso, Apr 26 '23 at 11:09
Notice that $\sin\left(x+\frac{\pi}{2}\right)$ is an even function while $\cos\left(x+\frac{\pi}{2}\right)$ is an odd function; raising the former to a natural power and the latter to an even power gives even functions; the product of two even functions gives an even function; so $\sin^5\left(x+\frac{\pi}{2}\right)\cos^4\left(x+\frac{\pi}{2}\right)$ is an even function; so, $x=\frac{\pi}2$ is a line of symmetry of $\sin^5\left(x\right)\cos^4\left(x\right).$ — ryang, Apr 26 '23 at 12:18
@ryang Hi sensei :-) Isn't $x=0$ supposed to be the line of symmetry for even functions? What I mean is, an even function necessarily only has one line of symmetry ($x=0$). So, does $\sin^5\theta\cos^4\theta$ being an even function have importance to it being symmetric about $x= \frac{\pi}{2}$? — tryingtobeastoic, Apr 26 '23 at 18:03
Where did you see the claim even functions only have one line of symmetry? In contrast, it's true that if a function has more than one vertical line of symmetry, it's periodic; and if a periodic function has at least one vertical line of symmetry, then it has infinitely many. This happens all the time with the standard trig functions. — Brian Moehring, Apr 26 '23 at 18:28
Hey tryingtobeastoic. Any function $f$ such that $f(x)=f(-x)$ is, by definition, an even function; so, for example, $\cos$ is even; and, proving Brian's above assertions involves invoking this property. P.S. That full-stop that you edited away was actually correct to be there! — ryang, Apr 27 '23 at 04:48
@BrianMoehring No, you misunderstood. I meant that even functions must have at least one line of symmetry at $x=0$ (if the function is a function of x). "and if a periodic function has at least one vertical line of symmetry, then it has infinitely many". Thank you, I did not know that. — tryingtobeastoic, Apr 27 '23 at 07:52
It might be also mentioned that these symmetries get us the "communtation" of exponents $$ 2\int_0^{\frac{\pi}{2}}\sin^5(\theta)\cos^4(\theta) \ \ d\theta\ \ \ = \ \ 2\int_0^{\frac{\pi}{2}}\sin^4(\theta)\cos^5(\theta) \ \ d\theta\ $$ $$ = \ \frac{\Gamma\left( 3 \right)\cdot\Gamma\left(\frac52\right)}{ \Gamma\left( \frac{11}{2}\right)} \ \ = \ \ \mathbf{B} \left(3 \ , \ \frac52 \right) \ \ = \ \ \frac{\Gamma\left(\frac52 \right) · \Gamma\left( 3 \right) }{ \Gamma\left( \frac{11}{2}\right)} \ \ = \ \ \mathbf{B} \left(\frac52 \ , \ 3 \right) \ \ . $$ — , Apr 27 '23 at 08:28

score 3 · Accepted Answer · edited Apr 26 '23 at 18:11

3

We know that $$\int_0^{2a} f(x)dx=2\int_0^af(x)dx$$

if $f(a+x)=f(a-x)$
here $f(\theta)=\sin^5(\theta)\cos^4\theta$

Check $f(\pi/2+\theta)=f(\pi/2-\theta)$.

So, $$\int_0^{{2\cdot\frac{\pi}{2}}}\sin^5(\theta)\cos^4(\theta)d\theta=2\int_0^{\pi/2}\sin^5(\theta)\cos^4(\theta)d\theta$$

Rest is just application of Walli's formula.

edited Apr 26 '23 at 18:11

tryingtobeastoic

3,176

answered Apr 26 '23 at 12:07

RRR

376

2

https://math.stackexchange.com/questions/2833731/reduction-formula-for-integral-sinm-x-cosn-x-with-limits-0-to-pi-2 , here's the link to refer the Wallis formula – NadiKeUssPar Apr 26 '23 at 16:28

score 1 · Answer 2 · answered Apr 27 '23 at 09:55

Alternative explanation: $$\begin{align}\int_{\frac\pi 2}^{\pi}\sin^5\theta\cos^4\theta d\theta &\overset{\theta\rightarrow\theta+\frac\pi 2}{=} \int_0^{\frac{\pi}2}\sin^5(\theta+\frac\pi 2)\cos^4(\theta+\frac\pi 2)d(\theta+\frac\pi 2)\\ &\overset{\theta\rightarrow\frac\pi 2-\theta}{=}\int_0^{\frac{\pi}2}\cos^5\theta\sin^4\theta d\theta\\ &=\int_0^{\frac{\pi}2}\sin^5\theta\cos^4\theta d\theta \end{align}$$

$\displaystyle\int_0^\pi\sin^5\theta\cos^4\theta \,\mathrm d\theta=2\int_0^{\frac{\pi}{2}}\sin^5\theta\cos^4\theta \,\mathrm d\theta$

2 Answers2