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Some definite integrals look impossible until a mirror is used…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Some definite integrals look impossible until a mirror is used. The trick in this problem works whenever the interval of integration is symmetric about $0$ and the integrand is built in a special way, from one even function and one odd function. Let $f$ be continuous and even on $[-a, a]$ , so that $f(-x) = f(x)$ , and let $g$ be continuous and odd there, so that $g(-x) = -g(x)$ . The figure shows the idea for $f(x) = x^2$ and $g(x) = x\ln 3$ on $[-2, 2]$ . It shows the curve $y = \dfrac{x^2}{1 + 3^x}$ , its mirror image in the $y$ -axis, and their sum, which is simply $y = x^2$ . The two curves are very different on their own, yet the area under each of them is the same, and together they make the area under the parabola.

Three curves on the interval from x = -2 to x = 2. A solid curve y = x^2/(1 + 3^x) is high on the left and low on the right. A dashed curve, its mirror image in the y-axis, is low on the left and high on the right. A dotted parabola y = x^2 lies above both and is their sum at every x.

Prove that

\displaystyle \int_{-a}^{a} \frac{f(x)}{1 + e^{g(x)}}\,dx = \int_0^{a} f(x)\,dx.

Write the proof in full.

Then use (a), with the right choice of $f$ and $g$ each time, to find these integrals. Say what $f$ and $g$ are, and check that $f$ is even and $g$ is odd.

$\displaystyle\int_{-2}^{2} \frac{x^2}{1 + 3^x}\,dx$
$\displaystyle\int_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{1 + e^{\sin x}}\,dx$
$\displaystyle\int_{-3\pi}^{3\pi} \frac{|\sin x|}{1 + 2^x}\,dx$
$\displaystyle\int_{-\pi}^{\pi} \frac{x\sin x}{1 + e^x}\,dx$

Give each value to two decimals. None of these integrands has an antiderivative that can be written down by the usual rules, so a direct attack will not work.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

(a) Proof. Call the left side $I$ . Put $x = -u$ , so $dx = -du$ , and the limits $-a$ , $a$ become $a$ , $-a$ . Then $I = \displaystyle\int_{-a}^{a} \frac{f(-u)}{1 + e^{g(-u)}}\,du = \int_{-a}^{a} \frac{f(u)}{1 + e^{-g(u)}}\,du = \int_{-a}^{a} \frac{f(u)\,e^{g(u)}}{e^{g(u)} + 1}\,du.$ Adding this to the first form of $I$ : $2I = \displaystyle\int_{-a}^{a} f(x)\,\frac{1 + e^{g(x)}}{1 + e^{g(x)}}\,dx = \int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx$ , because $f$ is even. Dividing by $2$ proves the result. Nothing else about $f$ and $g$ is needed.

(b) $f(x) = x^2$ (even), $g(x) = x\ln 3$ (odd): the integral is $\displaystyle\int_0^2 x^2\,dx = \frac83 \approx 2.67$ .

(c) $f(x) = \cos^2 x$ (even), $g(x) = \sin x$ (odd): $\displaystyle\int_0^{\pi/2} \cos^2 x\,dx = \frac{\pi}{4} \approx 0.79$ .

(d) $f(x) = |\sin x|$ (even), $g(x) = x\ln 2$ (odd): $\displaystyle\int_0^{3\pi} |\sin x|\,dx = 3\int_0^{\pi} \sin x\,dx = 3 \times 2 = 6.00$ , since $|\sin x|$ has period $\pi$ .

(e) $f(x) = x\sin x$ , which is even because both factors are odd; $g(x) = x$ . By parts, $\displaystyle\int_0^{\pi} x\sin x\,dx = \Big[-x\cos x\Big]_0^{\pi} + \int_0^{\pi} \cos x\,dx = \pi + 0 = \pi \approx 3.14$ .

Check. Numerical integration of the four original integrands, over the whole intervals, gives the same four values to twelve decimals.

Watch out. The rule halves the interval, not the integrand. Integrating $\dfrac{x^2}{1 + 3^x}$ itself from $0$ to $2$ gives only about $0.45$ . And the rule needs $f$ even: in (e), $x\sin x$ qualifies only because it is a product of two odd functions.

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