Definite Integral of x over a squared cos squared x plus b squared sin squared x

This lesson evaluates the definite integral

$I = \int_0^{\pi} \frac{x\,dx}{a^2\cos^2 x + b^2\sin^2 x}.$

It is the last question of the miscellaneous exercise and a frequently asked high-mark question. The idea is to first remove the $x$ in the numerator using a symmetry property, then reduce the limits and finish with two standard substitutions.

Removing the x with the king property

Apply $\displaystyle\int_0^{a} f(x)\,dx = \int_0^{a} f(a-x)\,dx$ with $a = \pi$. Since $\cos^2(\pi - x) = \cos^2 x$ and $\sin^2(\pi - x) = \sin^2 x$, the denominator is unchanged, so

$I = \int_0^{\pi} \frac{(\pi - x)\,dx}{a^2\cos^2 x + b^2\sin^2 x}.$

Split the numerator into $\pi$ and $-x$:

$I = \pi\int_0^{\pi} \frac{dx}{a^2\cos^2 x + b^2\sin^2 x} - I.$

The second term is the original integral $I$. Transposing it to the left gives

$2I = \pi\int_0^{\pi} \frac{dx}{a^2\cos^2 x + b^2\sin^2 x}.$

Halving the limits

Let $f(x) = \dfrac{1}{a^2\cos^2 x + b^2\sin^2 x}$. Because $f(\pi - x) = f(x)$, the property $\displaystyle\int_0^{2a} f(x)\,dx = 2\int_0^{a} f(x)\,dx$ applies (here $2a = \pi$, so $a = \tfrac{\pi}{2}$):

$2I = \pi \cdot 2\int_0^{\pi/2} \frac{dx}{a^2\cos^2 x + b^2\sin^2 x},$

so

$I = \pi\int_0^{\pi/2} \frac{dx}{a^2\cos^2 x + b^2\sin^2 x}.$

Splitting at pi over 4

Use $\displaystyle\int_a^{b} = \int_a^{c} + \int_c^{b}$ with $c = \tfrac{\pi}{4}$, writing $I = I_1 + I_2$ where

$I_1 = \pi\int_0^{\pi/4} \frac{dx}{a^2\cos^2 x + b^2\sin^2 x}, \qquad I_2 = \pi\int_{\pi/4}^{\pi/2} \frac{dx}{a^2\cos^2 x + b^2\sin^2 x}.$

First piece: substitute tan x

Divide numerator and denominator of $I_1$ by $\cos^2 x$:

$I_1 = \pi\int_0^{\pi/4} \frac{\sec^2 x\,dx}{a^2 + b^2\tan^2 x}.$

Put $t = \tan x$, so $dt = \sec^2 x\,dx$. The limits become $t = 0$ at $x = 0$ and $t = 1$ at $x = \tfrac{\pi}{4}$:

$I_1 = \pi\int_0^{1} \frac{dt}{a^2 + b^2 t^2} = \frac{\pi}{b^2}\int_0^{1} \frac{dt}{\left(\tfrac{a}{b}\right)^2 + t^2}.$

Using $\displaystyle\int \frac{dt}{p^2 + t^2} = \tfrac{1}{p}\tan^{-1}\tfrac{t}{p}$ with $p = \tfrac{a}{b}$:

$I_1 = \frac{\pi}{b^2}\cdot\frac{b}{a}\Big[\tan^{-1}\tfrac{bt}{a}\Big]_0^{1} = \frac{\pi}{ab}\tan^{-1}\frac{b}{a}.$

Second piece: substitute cot x

For $I_2$, divide numerator and denominator by $\sin^2 x$:

$I_2 = \pi\int_{\pi/4}^{\pi/2} \frac{\csc^2 x\,dx}{a^2\cot^2 x + b^2}.$

Put $u = \cot x$, so $-\csc^2 x\,dx = du$. The limits become $u = 1$ at $x = \tfrac{\pi}{4}$ and $u = 0$ at $x = \tfrac{\pi}{2}$. The minus sign and the swap of limits cancel:

$I_2 = \pi\int_0^{1} \frac{du}{a^2 u^2 + b^2} = \frac{\pi}{a^2}\int_0^{1} \frac{du}{u^2 + \left(\tfrac{b}{a}\right)^2}.$

With $p = \tfrac{b}{a}$:

$I_2 = \frac{\pi}{a^2}\cdot\frac{a}{b}\Big[\tan^{-1}\tfrac{au}{b}\Big]_0^{1} = \frac{\pi}{ab}\tan^{-1}\frac{a}{b}.$

Combining the pieces

Add the two results:

$I = I_1 + I_2 = \frac{\pi}{ab}\left(\tan^{-1}\frac{b}{a} + \tan^{-1}\frac{a}{b}\right).$

Since $\tan^{-1}\tfrac{b}{a} = \cot^{-1}\tfrac{a}{b}$, this becomes

$I = \frac{\pi}{ab}\left(\cot^{-1}\frac{a}{b} + \tan^{-1}\frac{a}{b}\right).$

Using $\tan^{-1}\theta + \cot^{-1}\theta = \tfrac{\pi}{2}$:

$I = \frac{\pi}{ab}\cdot\frac{\pi}{2} = \frac{\pi^2}{2ab}.$

Key takeaways

• Replacing $x$ with $\pi - x$ turns the $x$ in the numerator into a copy of $I$, which you transpose to solve for $I$.
• Because the remaining integrand is even about $\tfrac{\pi}{2}$, the limits halve to $0$ to $\tfrac{\pi}{2}$.
• Splitting at $\tfrac{\pi}{4}$ and substituting $\tan x$ and $\cot x$ gives two inverse-tangent terms whose sum collapses to $\dfrac{\pi^2}{2ab}$.