Q.$\int_{0}^{\pi} x\log\sin x\,dx$

The key idea is that when you have an integrand of the form $\frac{a\sin x + b\cos x}{c\sin x + d\cos x}$, you can often rewrite the numerator as $A(\text{denominator}) + B(\text{derivative of denominator})$. This works because the derivative of $\sin x + \cos x$ is $\cos x - \sin x$, and any linear combination of $\sin x$ and $\cos x$ can be expressed as a combination of these two. Once you do that, the integral splits into a simple constant term and a logarithmic term that often vanishes over symmetric limits.

1. Set up the decomposition We want constants $A$ and $B$ such that

Expanding the right side:

Matching coefficients with $3\sin x + 4\cos x$ gives the system:

Adding the equations: $2A = 7 \Rightarrow A = \frac{7}{2}$.

Subtracting: $2B = 1 \Rightarrow B = \frac{1}{2}$.

2. Rewrite the integral The integrand becomes:

So the integral is:

3. Evaluate the first part

4. Evaluate the second part Notice that the numerator $\cos x - \sin x$ is exactly the derivative of the denominator $\sin x + \cos x$. So let $u = \sin x + \cos x$, then $du = (\cos x - \sin x)\,dx$. When $x = 0$, $u = \sin 0 + \cos 0 = 1$. …

We are asked to compute

Concept and intuition:

When an integral over a symmetric interval (here $[0, \pi/2]$ is symmetric about $\pi/4$) contains a function that changes sign under a reflection of the variable, the integral often cancels to zero. Specifically, if we let $x \to \frac{\pi}{2} - x$, then $\sin x$ becomes $\cos x$ and vice versa. This swaps the numerator and denominator inside the log, flipping the sign of the integrand. Over the same interval, the positive and negative contributions exactly cancel.

Let’s verify this step by step.

1. Apply the substitution $x \to \frac{\pi}{2} - x$. Let $t = \frac{\pi}{2} - x$. Then $dx = -dt$, and when $x = 0$, $t = \frac{\pi}{2}$; when $x = \frac{\pi}{2}$, $t = 0$. So

2. Simplify the logarithm. Notice that

Therefore,

3. Solve for $I$. The equation $I = -I$ implies $2I = 0$, so $I = 0$. …

Concept & Intuition

When an integrand looks messy but involves a function of $\cos x$, a classic trick is to exploit the symmetry $\cos(\pi - x) = -\cos x$. This often turns the integral into something like $\int_0^\pi \frac{1}{2}\,dx$, because the sum of the integrand at $x$ and at $\pi - x$ simplifies to a constant. Here, the denominator $e^{\cos x} + e^{-\cos x}$ is symmetric in a way that makes this work perfectly.

Step-by-step solution

1. Define the integral Let

2. Use the substitution $x \to \pi - x$ Replace $x$ by $\pi - x$. Then $dx$ becomes $-dx$, but the limits swap: when $x=0$, $\pi - x = \pi$; when $x=\pi$, $\pi - x = 0$. So

3. Simplify using $\cos(\pi - x) = -\cos x$ Then $e^{\cos(\pi - x)} = e^{-\cos x}$ and $e^{-\cos(\pi - x)} = e^{\cos x}$. Hence

4. Add the two expressions for $I$ We have

Adding them:

Step 1 — Name the integral and state the property

Let $a = 4042$ and

The king property of definite integrals says

(It follows from the substitution $u = a - x$: the limits swap, and $dx = -du$ restores the orientation. It is the standard weapon whenever the integrand shows a symmetric pairing like $\sqrt x$ against $\sqrt{a-x}$ — as here.)

Step 2 — Write the reflected copy

Replace $x \to a - x$ in the integrand. Then $\sqrt x \to \sqrt{a-x}$ and $\sqrt{a - x} \to \sqrt{x}$:

The denominator is unchanged (it is symmetric), and only the numerator has flipped. That is exactly why this trick works.

Step 3 — Add (1) and (2) …