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:


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?

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

2 Answers 2


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.

  • $\begingroup$ 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 $\int_{0}^{0}f(u)\,du \stackrel{?}{=} 0$. $\endgroup$
    – kennysong
    Mar 28, 2013 at 19:43
  • $\begingroup$ @copper.hat, In your first big equation don't you need $f'(x)$ on the LHS? $\endgroup$
    – Cantor
    Mar 28, 2013 at 20:05
  • $\begingroup$ @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). $\endgroup$
    – copper.hat
    Mar 28, 2013 at 20:47
  • $\begingroup$ @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. $\endgroup$
    – copper.hat
    Mar 28, 2013 at 21:04

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

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

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


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