Q.Evaluate $\int_0^{\pi} \dfrac{x \sin x}{1 + \cos^2 x}\, dx$

The key idea is to use the symmetry property of definite integrals: $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$.

Let $I = \int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x}\, dx$.

Replace $x$ by $\pi - x$:

$I = \int_0^{\pi} \frac{(\pi - x) \sin(\pi - x)}{1 + \cos^2(\pi - x)}\, dx = \int_0^{\pi} \frac{(\pi - x) \sin x}{1 + \cos^2 x}\, dx$.

Add the two expressions for $I$:

$2I = \int_0^{\pi} \frac{\pi \sin x}{1 + \cos^2 x}\, dx$.

Now evaluate the simpler integral. Let $u = \cos x$, so $du = -\sin x\, dx$. When $x=0$, $u=1$; when $x=\pi$, $u=-1$.

$2I = \pi \int_1^{-1} \frac{-du}{1+u^2} = \pi \int_{-1}^{1} \frac{du}{1+u^2} = \pi \left[ \tan^{-1} u \right]_{-1}^{1} = \pi \left( \frac{\pi}{4} - (-\frac{\pi}{4}) \right) = \pi \cdot \frac{\pi}{2} = \frac{\pi^2}{2}$.

Thus $I = \frac{\pi^2}{4}$.

When you see an integral from $0$ to $\pi$ (or $0$ to $a$) with a product of $x$ and a trigonometric function, the first instinct should be to check if symmetry can help. The standard trick is to use the property:

This works because replacing $x$ by $a-x$ just reverses the order of integration, but the limits stay the same. Here $a = \pi$, so we replace $x$ by $\pi - x$ in the integrand.

Let’s see what happens.

1. Apply the symmetry property

Let

Using $x \to \pi - x$, we get:

Now recall: $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$, so $\cos^2(\pi - x) = \cos^2 x$. Therefore:

2. Add the two expressions for $I$

We now have two expressions for the same $I$:

Add them:

So:

The $x$ has vanished — that’s the whole point. Now we just need to evaluate the remaining trigonometric integral.

3. Evaluate the trigonometric integral

Let $J = \int_0^{\pi} \frac{\sin x}{1 + \cos^2 x}\, dx$.

Substitute $u = \cos x$, so $du = -\sin x\, dx$. When $x = 0$, $u = 1$; when $x = \pi$, $u = -1$. Thus:

This is a standard integral:

So:

4. Finish solving for $I$

We have $2I = \pi \cdot J = \pi \cdot \frac{\pi}{2} = \frac{\pi^2}{2}$.

Therefore:

Method: The Symmetry Property $\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$

Use this for a definite integral over $[0,a]$ whose integrand simplifies when $x$ is replaced by $a-x$, especially when an isolated $x$ factor multiplies a symmetric function.

Steps

Step 1: Write $I$ and form the reflected copy.

Let $I=\int_0^a f(x)\,dx$ and also $I=\int_0^a f(a-x)\,dx$. Replacing $x\to a-x$ leaves $\sin,\cos^2$ etc. unchanged but turns the isolated $x$ into $a-x$.

Step 2: Add the two forms.

Adding gives $2I=\int_0^a\big[f(x)+f(a-x)\big]\,dx$, in which the $x$-factor combines into the constant $a$, cancelling the troublesome $x$.

Step 3: Evaluate the simpler integral, then divide by 2.

For $\int_0^\pi\frac{x\sin x}{1+\cos^2x}\,dx$, this reduces to $\frac\pi2\int_0^\pi\frac{\sin x}{1+\cos^2x}\,dx$; a substitution $u=\cos x$ then gives $\frac{\pi^2}{4}$.

Common Mistakes

Mistake 1: Assuming $f(a-x)=f(x)$ for every term.

Why it's wrong: only $\sin x$ and $\cos^2 x$ are unchanged under $x\to\pi-x$; the isolated $x$ becomes $\pi-x$. Correct approach: substitute carefully into each factor.

Mistake 2: Forgetting to divide by 2 at the end.

Why it's wrong: adding the two forms gives $2I$, so the final answer needs the factor $\frac12$. Correct approach: solve $2I=\cdots$ for $I$.

Mistake 3: Mishandling the leftover substitution.

Why it's wrong: $\int_0^\pi\frac{\sin x}{1+\cos^2x}\,dx$ needs $u=\cos x$ with limits $1\to-1$; a sign slip mis-evaluates the arctan. Correct approach: change limits and track the minus from $du=-\sin x\,dx$.