Question

Evaluate: \(\displaystyle\int_0^\pi \frac{x\,dx}{a^2\cos^2x+b^2\sin^2x}\)

Hint:

Use ∫f(x)dx=∫f(a−x)dx over [0,π] to remove x, then evaluate the resulting symmetric integral.

Answer 1

Step 1: Key Formula or Approach:
Use the property \(\displaystyle\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx\). Let \(I\) denote the given integral and apply this with \(a=\pi\).

Step 2: Applying the property:
Let \(I = \displaystyle\int_0^\pi \frac{x\,dx}{a^2\cos^2x+b^2\sin^2x}\). Replacing \(x\) by \(\pi-x\): since \(\cos(\pi-x)=-\cos x\) and \(\sin(\pi-x)=\sin x\), and both appear squared, the denominator is unchanged:
\[ I = \int_0^\pi \frac{(\pi-x)\,dx}{a^2\cos^2x+b^2\sin^2x} \]

Step 3: Adding the two forms of \(I\):
\[ 2I = \int_0^\pi\frac{x\,dx}{a^2\cos^2x+b^2\sin^2x} + \int_0^\pi\frac{(\pi-x)\,dx}{a^2\cos^2x+b^2\sin^2x} = \pi\int_0^\pi\frac{dx}{a^2\cos^2x+b^2\sin^2x} \]

Step 4: Evaluating \(J=\displaystyle\int_0^\pi\frac{dx}{a^2\cos^2x+b^2\sin^2x}\):
By the symmetry of \(\cos^2x\) and \(\sin^2x\) about \(x=\pi/2\), \(\displaystyle\int_0^\pi = 2\int_0^{\pi/2}\). Using the standard result \(\displaystyle\int_0^{\pi/2}\frac{dx}{a^2\cos^2x+b^2\sin^2x} = \frac{\pi}{2ab}\) (divide numerator and denominator by \(\cos^2x\), substitute \(t=\tan x\)), we get \(J = 2\cdot\dfrac{\pi}{2ab} = \dfrac{\pi}{ab}\).

Step 5: Solving for \(I\):
\[ 2I = \pi\cdot\frac{\pi}{ab} = \frac{\pi^2}{ab} \implies I = \frac{\pi^2}{2ab} \]

Final Answer:
\[ I = \frac{\pi^2}{2ab} \] \[ \boxed{\dfrac{\pi^2}{2ab}} \]