3
$\begingroup$

Evaluate $$\int_0^\pi \sin^4\theta \cos\theta\, d\theta.$$

This is clearly very simple under indefinite circumstances, that is, to substitute $u=\sin\theta$, and work from there, but that leaves me with $0-0$, which is wrong, certainly.

$\endgroup$
  • 6
    $\begingroup$ Why should it be wrong? $\endgroup$ – Tim Raczkowski Mar 7 '15 at 23:39
  • $\begingroup$ Because there is physical space in between them. $\endgroup$ – Yaniv Proselkov Mar 7 '15 at 23:41
  • 1
    $\begingroup$ The integrand is "odd symmetric" about $\pi/2$ so 0 is the answer. $\endgroup$ – jdods Mar 7 '15 at 23:41
  • $\begingroup$ What exactly does "them" mean in that comment? $\endgroup$ – Old John Mar 7 '15 at 23:42
  • 2
    $\begingroup$ Physical space can be quantified both negatively and positively depending on what structures are imposed on that physical space. Vast amounts of physical space can be "zero". $\endgroup$ – jdods Mar 7 '15 at 23:44
7
$\begingroup$

It is not wrong. The indefinite integral evaluates to $$\sin^5(\theta)/5$$ so when you take $\sin^5(\pi)/5 - \sin^5(0)/5$ you get $0-0$ like you said.

$\endgroup$
3
$\begingroup$

You don't need substitution you can do $$ \frac{d}{d\theta}\sin^5\theta = 5 \sin^4\theta \cos \theta $$ This your integral is $$ \int \frac{1}{5}\frac{d}{d\theta}\sin^5\theta d\theta $$ Which means you just apply the boundary.

$\endgroup$
  • $\begingroup$ Isn't that just the same as substituion?? $\endgroup$ – user99914 Mar 8 '15 at 0:09
  • $\begingroup$ @john You could think it that way, but since the derivative is only in terms of $\theta$ then you can perform the integral straight away without changing the limits. Also, since you have a multi-valued substitution i.e. $(0,\pi)\to(0,0)$ then using the above is clearer. But it is just my humble opinion of an approach :). $\endgroup$ – Chinny84 Mar 8 '15 at 0:18
1
$\begingroup$

No need to substitute $u=\sin\theta$.

$$\int_0^{\pi}\sin^4\theta\cos\theta\,d\theta=\int_0^{\pi/2}\sin^4\theta\cos\theta\,d\theta+\int_0^{\pi/2}\sin^4(\pi-\theta)\cos(\pi-\theta)\,d\theta$$

$$=\int_0^{\pi/2}\sin^4\theta\cos\theta\,d\theta-\int_0^{\pi/2}\sin^4\theta\cos\theta\,d\theta=0$$

$\endgroup$
1
$\begingroup$

i don't think you even need to evaluate this integral. just the symmetry of $\sin$ and $\cos$ would do. that is we will use the facts $$\sin(\pi - t) = \sin t, \cos(\pi - t) = -\cos t.$$

we will make a change of variable $$u = \pi - \theta, \theta = \pi - u, du = -d \theta$$ so that $$I = \int_0^\pi \sin^4 \theta \, \cos \theta \, d \theta = \int_\pi^0\sin^4 u(-\cos u)(-d u)= -\int_0^\pi\sin^4 u \,cos u \,du =-I$$

so that $$2I = 0 \to I = 0. $$

$\endgroup$
  • $\begingroup$ This is the nicest answer! One can present it more conceptually too. $\sin x$ is symmetric around $\pi/2$: from $\sin 0 = 0$, it goes up to $\sin \pi/2 = 1$, then down to $\sin \pi = 0$. So $\sin^4 x$ is also symmetric. But $\cos x$ is antisymmetric around $\pi/2$: it goes from $\cos 0 = 1$ to $\cos \pi/2 = 0$, and then on down to $\cos \pi = -1$, mirroring in negative what it did between $0$ and $\pi/2$. So $\sin^4 x \cos x$ is also antisymmetric; so integrating it, its (negative) area between $x= \pi/2$ and $x=\pi$ exactly cancels out its positive area between $x=0$ and $x= \pi/2$. $\endgroup$ – Peter LeFanu Lumsdaine Mar 14 '15 at 12:26
  • 1
    $\begingroup$ @PeterLeFanuLumsdaine, thanks. i appreciate it. my students are not too fond of the geometric arguments. i hope they will see the benefits of this later. $\endgroup$ – abel Mar 14 '15 at 12:29
0
$\begingroup$

If you are familiar with linear algebra, you can use the inner product to evaluate the integral. Since the $\sin^4(\theta)\cos(\theta)$ is even, $$ \int_0^{\pi}\sin^4(\theta)\cos(\theta)d\theta = \frac{1}{2}\int_0^{2\pi}\sin^4(\theta)\cos(\theta)d\theta\tag{1} $$ The orthonormal basis for the functions $\sin(\theta)$ and $\cos(\theta)$ are $$ \bigl\{1/\sqrt{2},\cos(n\theta),\sin(n\theta)\bigr\} $$ where $n\in\mathbb{Z}$. The inner product of this space is $$ \langle f,g\rangle = \frac{1}{\pi}\int_0^{2\pi}fg \ d\theta=\delta(f,g)= \begin{cases} 1, & f=g\\ 0, & f\neq g \end{cases} $$ We can write equation $(1)$ as \begin{align} \frac{1}{2}\int_0^{2\pi}\sin^4(\theta)\cos(\theta)d\theta & = \frac{\pi}{2}\biggl[\frac{1}{\pi}\int_0^{2\pi}\sin^4(\theta)\cos(\theta)d\theta\biggr]\\ &= \frac{\pi}{2}\biggl[\frac{1}{\pi}\int_0^{2\pi}\frac{1}{2\sqrt{2}}\frac{1}{\sqrt{2}}\cos(\theta)d\theta - \frac{1}{\pi}\int_0^{2\pi}\frac{1}{2}\cos(\theta)\cos(2\theta)d\theta\\ &+\frac{1}{\pi}\int_0^{2\pi}\frac{1}{4\sqrt{2}}\frac{1}{\sqrt{2}}\cos(\theta)d\theta + \frac{1}{\pi}\int_0^{2\pi}\frac{1}{8}\cos(\theta)\cos(4\theta)d\theta\biggr]\\ & = \frac{\pi}{4\sqrt{2}}\langle 1/\sqrt{2},\cos(\theta)\rangle - \frac{\pi}{4}\langle\cos(\theta),\cos(2\theta)\rangle + \frac{\pi}{8\sqrt{2}}\langle 1/\sqrt{2},\cos(\theta)\rangle\\ & + \frac{\pi}{16}\langle\cos(\theta),\cos(4\theta)\rangle \end{align} Since every inner product consists of $f\neq g$, every inner product is a constant times $\langle \ , \ \rangle = 0$; therefore, the integral is zero.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.