Example 30: $\int_0^{\pi}\frac{x\sin x}{1+\cos^2x}\,dx$

Basics you need

Solution

1. Continuity, and why a direct attack fails

   Since $\cos^2 x \ge 0$⁠, the denominator $1 + \cos^2 x \ge 1$ is never 0, so the integrand is continuous on $[0, \pi]$ and $I$ is a definite number.

   Finding an antiderivative directly is not practical: integrating by parts with $u = x$ and $v = -\tan^{-1}(\cos x)$ leads to $\int\tan^{-1}(\cos x)\,dx$⁠, whose antiderivative none of the methods of this chapter can find. (Over $[0, \pi]$ that last integral is in fact 0, because $\tan^{-1}\cos(\pi - x)$ equals $-\tan^{-1}\cos x$⁠; but seeing this needs property P4 anyway.) So we use a property of definite integrals from the start.

   Properties P0 to P7 can give the value of a definite integral even when no usable antiderivative is available. That is exactly the situation here.
   A quotient of continuous functions is continuous where the denominator is not 0. Lesson: definite integrals

2. Apply P4 with $a = \pi$

   Replace every $x$ by $\pi - x$⁠:

   $$\begin{aligned}I &= \int_0^{\pi}\frac{(\pi - x)\sin(\pi - x)}{1 + \cos^2(\pi - x)}\,dx \\ &= \int_0^{\pi}\frac{(\pi - x)\sin x}{1 + \cos^2 x}\,dx.\end{aligned}$$
   P4 needs nothing except continuity of the integrand on $[0, a]$⁠. The two identities $\sin(\pi - x) = \sin x$ and $\cos^2(\pi - x) = (-\cos x)^2 = \cos^2 x$ leave everything unchanged except the factor $x$⁠, which becomes $\pi - x$⁠. That is the whole point of the choice $a = \pi$⁠.
   P4, Section 7.10 (a particular case of P3 with lower limit 0). Lesson: properties of definite integrals

3. Split and solve for $I$

   Write $\pi - x$ as two terms and recognise $I$ again:

   $$\begin{aligned}I &= \pi\int_0^{\pi}\frac{\sin x}{1 + \cos^2 x}\,dx \\ &\quad - \int_0^{\pi}\frac{x\sin x}{1 + \cos^2 x}\,dx \\ &= \pi J - I,\end{aligned}$$

   where $J = \int_0^{\pi}\frac{\sin x}{1 + \cos^2 x}\,dx$⁠. So $2I = \pi J$⁠, that is

   $$I = \frac{\pi}{2}\int_0^{\pi}\frac{\sin x}{1 + \cos^2 x}\,dx = \frac{\pi}{2}J.$$
   The second integral is the original $I$ word for word, so the equation $I = \pi J - I$ can be solved for $I$⁠. This is the same as adding the two forms of $I$ from Step 2: $x + (\pi - x) = \pi$⁠. The hard factor $x$ is gone.
   Linearity: $\int(\pi g - xg) = \pi\int g - \int xg$⁠. Moving $-I$ to the left gives $2I$⁠. Lesson: properties

4. Substitute $t = \cos x$ in $J$

   Then $dt = -\sin x\,dx$⁠. When $x = 0$⁠, $t = 1$⁠; when $x = \pi$⁠, $t = -1$⁠. So

   $$J = \int_1^{-1}\frac{-dt}{1 + t^2} = \int_{-1}^{1}\frac{dt}{1 + t^2}.$$
   $\sin x\,dx$ is (minus) the differential of $\cos x$⁠, and the rest is a function of $\cos x$⁠. The substitution is legitimate: $\cos x$ is differentiable and decreasing on $[0, \pi]$⁠, so $t$ runs once from 1 to $-1$⁠. The minus sign and the reversed limits cancel by P1 (swapping the limits changes the sign).
   P1, Section 7.10: $\int_1^{-1}(-g) = \int_{-1}^{1} g$⁠. Basics: substitution

5. Use evenness, then $\tan^{-1}$

   $g(t) = \frac{1}{1 + t^2}$ is even ($g(-t) = g(t)$⁠), so by P7 (i)

   $$J = 2\int_0^1\frac{dt}{1 + t^2} = 2\left[\tan^{-1}t\right]_0^1 = 2\left(\frac{\pi}{4} - 0\right) = \frac{\pi}{2}.$$

   Therefore

   $$I = \frac{\pi}{2}\cdot J = \frac{\pi}{2}\cdot\frac{\pi}{2} = \frac{\pi^2}{4} \approx 2.4674011.$$
   The book writes this as $I = \pi\int_0^1\frac{dt}{1+t^2} = \pi\left[\tan^{-1}t\right]_0^1 = \frac{\pi^2}{4}$⁠: the same numbers in a different order. One could also skip P7: $\left[\tan^{-1}t\right]_{-1}^{1} = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{2}$⁠.
   $\tan^{-1}1 = \frac{\pi}{4}$⁠, the principal value in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠. Basics: symmetry

6. The picture

   The integrand is lopsided (larger on the right, because of the factor $x$⁠). Averaging it with its reflection $x \mapsto \pi - x$ gives a symmetric curve with the same area, and that symmetric curve is easy to integrate.
   Area under a non-negative function = definite integral. Basics: area as a sum of thin strips

Every fact used, and where it comes from