Evaluate $3\int_{0}^{2\pi} \sin(t) \cos(t) \,{\rm d}t$

$$3\int_{0}^{2\pi} \sin(t) \cos(t) \,{\rm d}t$$

My calculus is a bit rusty and I can not find where I get it wrong. Setting $u= \sin(t)$, I get ${\rm d} u=\cos(t) \,{\rm d} t$ and, thus,

$$3\int_{u=0}^{u=0}u \,{\rm d}u=0$$

Solution 1:

The substitution is correct. If $f$ is a function with an antiderivative $F$, one has by the fundamental theorem of calculus \begin{equation} \int_a^b f(u(t))u'(t)dt = \int_a^b(F(u(t)))' dt = F(u(b)) - F(u(a)) = \int_{u(a)}^{u(b)}f(x) d x \end{equation} A sufficient hypothesis is that $u'$ is continuous in $[a,b]$ and that $f$ is continuous on an interval that contains $u([a,b])$.

In your case $u(t) = \sin(t)$ and $f(t)=t$.

I'm amazed that so many people are puzzled by this simple application of the fundamental theorem of calculus.

Let us take the case of the integral \begin{equation} I = \int_0^\pi \sin(x) d x \end{equation} and let $u(x) = - \cos(x)$ as suggested in the comments, with $f(x)= 1$. The above substitution formula gives \begin{equation} I = \int_{-\cos(0)}^{-\cos(\pi)} dx = \int_{-1}^1 d x = 2 \end{equation} which is the correct result.

There is no contradiction with this counterexample because in the counterexample, the invalid substitution is $u(x) = \sin(x)$. It would imply $f(u) = \pm\frac{u}{\sqrt{1-u^2}}$ and the original integral must be split at $\pi/2$ to choose between $+$ and $-$. It does not invalidate the above formula with the continuity condition on $f$.

Solution 2:

I thought I'd create a separate answer to just try and explain the confusion people have a bit more. @Gribouillis has given you a really good answer, so please give him the solution check-mark.

It's not true that the substitution needs to be injective. You can see for yourself on the Wikipedia article the conditions required for $u$ substitution, and injectivity is not one of those requirements: https://en.wikipedia.org/wiki/Integration_by_substitution. In this case - your solution for this integral is completely valid (it is indeed $0$) and there is nothing wrong with your method.

There are some cases where if you blindly apply substitution, you can get the incorrect result. For example, consider

$${\int_{0}^{\pi}\sin(x)dx}$$

Let's substitute ${u=\sin(x)}$. Then the bounds become ${\int_{0}^{0} ...du=0}$. Does that mean the original integral is $0$? NO!. We have incorrectly applied the theorem of Integration by Substitution. All substitution says is that

$${\int_{a}^{b}f(\phi(x))\phi'(x)dx=\int_{\phi(a)}^{\phi(b)}f(u)du}$$

${\int_{0}^{\pi}\sin(x)dx}$ does not match this form. You can make it happen - so let's try it. We can write

$${\int_{0}^{\pi}\sin(x)\frac{\cos(x)}{\cos(x)}dx}$$

Now here's the problem - we need to write the bottom ${\cos(x)}$ in terms of ${\sin(x)}$ - we know ${\cos(x)=\pm\sqrt{1-\sin^2(x)}}$ - that's the problem - the ${\pm}$. If ${x \in \left[0,\frac{\pi}{2}\right)}$, we have ${\cos(x)=\sqrt{1-\sin^2(x)}}$, and if ${x \in \left(\frac{\pi}{2},\pi\right]}$ then ${\cos(x)=-\sqrt{1-\sin^2(x)}}$. So you will have to split up the integral into two halves, like so:

$${\int_{0}^{\pi}\sin(x)dx=\int_{0}^{\frac{\pi}{2}}\frac{\sin(x)}{\sqrt{1-\sin^2(x)}}\cos(x)dx+\int_{\frac{\pi}{2}}^{\pi}\frac{\sin(x)}{-\sqrt{1-\sin^2(x)}}\cos(x)dx}$$

And now we can apply substitution to the two individual integrals to get

$${=\int_{0}^{1}\frac{u}{\sqrt{1-u^2}}du-\int_{1}^{0}\frac{u}{\sqrt{1-u^2}}du=2\int_{0}^{1}\frac{u}{\sqrt{1-u^2}}du}$$

So as you can see - there are certain subtleties with using substitution - but injectivity is not a direct condition required for it. Your solution to the integral is absolutely fine.