Q.Evaluate $\int_0^{\pi} \dfrac{x\, dx}{a^2 \cos^2 x + b^2 \sin^2 x}$

Use the King property $\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$ to remove the $x$ in the numerator.

Let $I=\displaystyle\int_0^{\pi}\frac{x\,dx}{a^2\cos^2 x+b^2\sin^2 x}$. Replacing $x\to\pi-x$ leaves the denominator unchanged (since $\cos^2(\pi-x)=\cos^2 x$, $\sin^2(\pi-x)=\sin^2 x$):

Adding the two forms:

Evaluate $J$. By symmetry about $x=\tfrac{\pi}{2}$, $J=2\displaystyle\int_0^{\pi/2}\frac{dx}{a^2\cos^2 x+b^2\sin^2 x}$. Divide by $\cos^2 x$ and put $t=\tan x$: …

The idea

The denominator $a^2\cos^2 x+b^2\sin^2 x$ is symmetric about $x=\tfrac{\pi}{2}$, but the numerator is just $x$. The substitution $x\to\pi-x$ is designed exactly to exploit that mismatch.

Step 1 — Apply the King property

Let

Using $\int_0^{a}f(x)\,dx=\int_0^{a}f(a-x)\,dx$ with $a=\pi$, replace $x$ by $\pi-x$. Since $\cos(\pi-x)=-\cos x$ and $\sin(\pi-x)=\sin x$, the denominator is unchanged:

Step 2 — Add the two forms

Adding the original and transformed integrals, the $x$ and $-x$ cancel:

Call the remaining integral $J$, so $2I=\pi J$.

Step 3 — Evaluate $J$

The integrand is symmetric about $x=\tfrac{\pi}{2}$, so

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

Method: King's property to kill the $x$, then a $\tan$ substitution

Use this for $\int_0^{\pi}\dfrac{x\cdot(\text{even-about-}\tfrac{\pi}{2}\text{ function})}{\cdots}\,dx$: the reflection property removes the linear $x$, reducing the problem to a pure trig-rational integral.

Steps

Step 1: Apply $x\to a-x$ (here $a=\pi$).

Because $\cos^2(\pi-x)=\cos^2 x$ and $\sin^2(\pi-x)=\sin^2 x$, the denominator is unchanged and

Step 2: Add the two forms to eliminate $x$.

$x+(\pi-x)=\pi$, so

Step 3: Evaluate the remaining $J$ by a $\tan$ substitution. …

Common Mistakes

Mistake 1: Thinking the substitution $x\to\pi-x$ changes the denominator.

Why it's wrong: although $\cos(\pi-x)=-\cos x$, the denominator uses $\cos^2 x$, so squaring removes the sign and it is unchanged — that invariance is exactly what makes the trick work. Correct approach: note the denominator depends only on $\cos^2 x,\sin^2 x$, both even under $x\to\pi-x$.

Mistake 2: Forgetting the $\tfrac12$ / limit handling when evaluating $J$. …