Question

Find the value of \(\displaystyle\int_0^{\pi}\dfrac{x\,dx}{a^2\cos^2x+b^2\sin^2x}\).

Hint:

Use \(x\to\pi-x\) to double the integral into a pure trig integral, then reduce to the standard \(\int_0^{\pi/2}\frac{dx}{a^2\cos^2x+b^2\sin^2x}=\frac{\pi}{2ab}\).

Answer 1

Step 1: Understanding the Concept:
Use the King's-rule property \(x\to\pi-x\) (since \(\cos^2\) and \(\sin^2\) are unchanged by \(x\to\pi-x\)) to remove the awkward factor of \(x\) from the numerator, reducing the problem to a purely trigonometric integral.

Step 2: Applying x -> pi - x:
Let \(I=\displaystyle\int_0^\pi\dfrac{x\,dx}{a^2\cos^2x+b^2\sin^2x}\). Since \(\cos^2(\pi-x)=\cos^2x\) and \(\sin^2(\pi-x)=\sin^2x\): \(I=\displaystyle\int_0^\pi\dfrac{(\pi-x)\,dx}{a^2\cos^2x+b^2\sin^2x}\).

Step 3: Adding the two forms:
\(2I=\pi\displaystyle\int_0^\pi\dfrac{dx}{a^2\cos^2x+b^2\sin^2x}\).

Step 4: Reducing the [0,pi] integral to [0,pi/2]:
The integrand has period \(\pi\) and is symmetric about \(x=\pi/2\) (check: replacing \(x\) by \(\pi-x\) leaves it unchanged), so \(\displaystyle\int_0^\pi(\cdots)dx=2\int_0^{\pi/2}(\cdots)dx\).

Step 5: Using the standard result for the quarter-period integral:
\(\displaystyle\int_0^{\pi/2}\dfrac{dx}{a^2\cos^2x+b^2\sin^2x}=\dfrac{\pi}{2ab}\) (dividing numerator and denominator by \(\cos^2x\), substituting \(t=\tan x\), and using \(\int_0^\infty\dfrac{dt}{a^2+b^2t^2}=\dfrac{\pi}{2ab}\)). So \(\displaystyle\int_0^\pi(\cdots)dx=2\times\dfrac{\pi}{2ab}=\dfrac{\pi}{ab}\).

Step 6: Solving for I:
\(2I=\pi\times\dfrac{\pi}{ab}=\dfrac{\pi^2}{ab}\ \Rightarrow\ I=\dfrac{\pi^2}{2ab}\).

Final Answer:
\(\displaystyle\int_0^\pi\dfrac{x\,dx}{a^2\cos^2x+b^2\sin^2x}=\boxed{\dfrac{\pi^2}{2ab}}\).