Definite Integral of $\sin^4(x)\cos(x).$

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.

• Why should it be wrong? – Tim Raczkowski Mar 7 '15 at 23:39
• Because there is physical space in between them. – Yaniv Proselkov Mar 7 '15 at 23:41
• The integrand is "odd symmetric" about $\pi/2$ so 0 is the answer. – jdods Mar 7 '15 at 23:41
• What exactly does "them" mean in that comment? – Old John Mar 7 '15 at 23:42
• 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". – jdods Mar 7 '15 at 23:44

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.

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.

• Isn't that just the same as substituion?? – user99914 Mar 8 '15 at 0:09
• @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 :). – Chinny84 Mar 8 '15 at 0:18

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

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

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

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

• 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 $x= \pi/2$. – Peter LeFanu Lumsdaine Mar 14 '15 at 12:26
• @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. – abel Mar 14 '15 at 12:29

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.