Animated Solution for Mathematics - Definite Integration: Evaluate: ${\int }_0^{\pi }\frac{xdx}{1+\cos \alpha \sin x},0<\alpha <\pi$

Welcome, fellow traveler on the path to JEE Advanced mastery. Today, we are not just solving a problem; we are witnessing a dance of symmetry and transformation. We are going to evaluate the integral:

where $0<\alpha <\pi$. At first glance, this integral looks intimidating. The $x$ in the numerator is a classic 'distractor'—it makes the standard antiderivative formulas impossible to apply directly.

But in the world of JEE calculus, whenever you see a variable $x$ in the numerator of a definite integral, your intuition should immediately scream: King's Property!

The King's Property Transformation

The King's Property, or the property of definite integrals, states that ${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx$. This is our secret weapon. By replacing $x$ with $(\pi -x)$, we are essentially 'flipping' the function.

Let's see what happens:

Recall your trigonometric identities: $\sin (\pi -x)=\sin x$. The denominator remains perfectly unchanged! This is the moment of magic. Our new integral is:

Now, watch what happens when we add our original integral $I$ to this new version. We get $2I$ on the left, and on the right, the $x$ and $-x$ in the numerators cancel out completely:

We have successfully banished the $x$ from the numerator. The problem is now reduced to a standard trigonometric integral.

Exploiting Symmetry

We can simplify this further. The function $f(x)=\frac{1}{1+\cos \alpha \sin x}$ is symmetric about $x=\pi /2$. Because of this, the integral from $0$ to $\pi$ is exactly twice the integral from $0$ to $\pi /2$.

Dividing both sides by $2$, we arrive at a much cleaner form:

The Half-Angle Substitution

Now, we enter the realm of algebraic substitution. To handle the $\sin x$ in the denominator, we use the Weierstrass substitution: $t=\tan (x/2)$. This implies $dx=\frac{2dt}{1+t^2}$ and $\sin x=\frac{2t}{1+t^2}$.

When $x=0$, $t=0$. When $x=\pi /2$, $t=\tan (\pi /4)=1$. Our limits are now $0$ to $1$. Substituting these into our integral:

Multiplying the numerator and denominator by $(1+t^2)$, we get:

Completing the Square

We are almost there. The denominator $t^2+2t\cos \alpha +1$ is a quadratic. To integrate this, we complete the square:

Our integral becomes:

This is the standard form $\int \frac{du}{u^2+a^2}=\frac{1}{a}{\tan }^{-1}(\frac{u}{a})$. Here, $u=t+\cos \alpha$ and $a=\sin \alpha$. Applying the formula:

The Final Trigonometric Simplification

Evaluating at the limits $1$ and $0$:

Using half-angle identities, $\frac{1+\cos \alpha }{\sin \alpha }=\frac{2{\cos }^2(\alpha /2)}{2\sin (\alpha /2)\cos (\alpha /2)}=\cot (\alpha /2)$. And $\frac{\cos \alpha }{\sin \alpha }=\cot \alpha$.

So we have:

Using ${\tan }^{-1}(\cot \theta )=\pi /2-\theta$, the expression becomes:

Simplifying the bracket: $(\pi /2-\alpha /2-\pi /2+\alpha )=\alpha /2$. Finally:

And there it is! The final result is: