# Making both bounds of integration zero

I came across a question while evaluating the integral:

$$\int_{0}^{\pi}\frac{\cos{t}}{1+9\sin^2{t}}\, dt$$

If you substitute $u=3\sin{t}$, you get:

$$\int_{0}^{0}\frac{1}{3+3u^2}\, du$$

Which evaluates to zero because(?) the bounds are both zero.

But then, can't you substitute any arbitrary expression to change both bounds to zero -- making the value zero? So in evaluating:

$$\int_{0}^{1}x\,dx$$

We could substitute $u = x^2-x$ or some trigonometric expression to change both bounds to zero. Clearly there is a misstep here, but which part of this substitution is invalid?

• The function may not be continuous. Mar 27, 2013 at 16:39
• The substitution should be monotone. Mar 27, 2013 at 16:40
• That's a bold statement. Mar 27, 2013 at 16:41
• No one has yet attempted to address the concerns in the last paragraph (about $u=x^2-x$).
– user641
Mar 27, 2013 at 19:46
• And I added a relevant comment to the answer below... Mar 27, 2013 at 20:05

There is no problem. See the comment by @BabyDragon above.

Note that the integrand satisfies $f(\frac{\pi}{2}-x) = -f(\frac{\pi}{2}+x)$ over the range of integration, so the answer is zero.

Regarding the last remark, note that for sufficiently smooth $f,u$, we have $$\int_{u(a)}^{u(b)} f(x) dx = \int_a^b f(u(t))u'(t) dt$$

With $a=0$, $b=1$, $f(x) = x$, and $u(t) = t^2-t$, this will result in

$$\int_0^0x dx = \int_0^1 (t^2-t)(2t-1)dt = 0$$

In particular, note how the change of variables affects the integration bounds.

• In your example, changing the bounds from zeroes to a numerical range preserves the value of the integral, but what if we change a range to zeroes (my original question)? i.e. given $\int_{0}^{1}x\,dx$ and $u = x^2-x$, we end up with $\int_{0}^{0}f(u)\,du \stackrel{?}{=} 0$. Mar 28, 2013 at 19:43
• @copper.hat, In your first big equation don't you need $f'(x)$ on the LHS? Mar 28, 2013 at 20:05
• @Cantor: No, its a change of variables. See any change of variable formula, eg, Theorem 7.26 in Rudin, "Real and complex analysis" (in particular the special case at the end). Mar 28, 2013 at 20:47
• @jellyksong: The above formula is the change of variables formula. When you use a substitution as above, the '$u$' is what appears in the original integral, eg, above would be $\int_0^\pi f(u(t))u'(t) dt$ ($u(t) = 3 \sin t$ here) and the equivalent integral would be $\int_{u(0)}^{u(\pi)} f(x)dx$. So, to use the formula as you have it, the original formula would have to have the form $\int_0^1 g(u(t))u'(t) dt$ for some $g$ with the given $u$. But it doesn't. Mar 28, 2013 at 21:04

An idea so that no substitution will be needed, based on $\displaystyle{\int\frac{f'}{1+f^2}=\arctan f}\;:$

$$\int\limits_0^\pi\frac{\cos t}{1+9\sin^2t}dt=\int\limits_0^\pi\frac{\cos t}{1+(3\sin t)^2}dt=$$

$$=\frac{1}{3}\int\limits_0^\pi\frac{(3\sin t)'dt}{1+(3\sin t)^2}=\left.\frac{1}{3}\arctan(3\sin t)\right|_0^\pi=\frac{1}{3}\left[\arctan 0-\arctan 0\right]=0$$