Q.By using the properties of definite integrals, evaluate the integral $\int_{0}^{\pi}\frac{x\,dx}{1+\sin x}$

The key idea is to use the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$, which exploits symmetry about the midpoint.

Let $I = \int_{0}^{\pi} \frac{x}{1+\sin x} \, dx$.

Applying the property with $a = \pi$:

Add the two expressions for $I$:

Now evaluate $J = \int_{0}^{\pi} \frac{dx}{1 + \sin x}$. Multiply numerator and denominator by $1 - \sin x$: …

Let’s look at the integral:

The presence of $x$ multiplied by a trigonometric function over a symmetric-looking interval $[0, \pi]$ is a classic signal. The trick is to use the property of definite integrals that exploits symmetry about the midpoint.

Why this property works

For any function $f(x)$ that is integrable on $[0, a]$, we have:

Why? Because substituting $t = a-x$ maps the interval $[0, a]$ onto itself in reverse. The area under the curve doesn’t change — it’s just a relabeling of the horizontal axis. This is especially powerful when $f(x)$ contains a factor like $x$, because $a-x$ introduces a complementary term that can simplify the sum.

Here, $a = \pi$, so we’ll replace $x$ by $\pi - x$ in the integrand.

Step-by-step solution

1. Apply the symmetry property

Let

Using $\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a-x)\,dx$ with $a = \pi$:

2. Simplify the trigonometric part

Recall that $\sin(\pi - x) = \sin x$. So the denominator stays the same:

Now we have two expressions for $I$:

3. Add the two expressions

Add them:

The $x$ terms cancel beautifully. So:

4. Evaluate the remaining integral …

Method: Reflection to cancel a linear factor $x$, then a rationalising trig trick

For $\displaystyle\int_0^a \frac{x}{h(x)}\,dx$ where $h(a-x)=h(x)$, reflecting turns $x$ into $a-x$; adding removes the $x$ and leaves a constant $\times$ a pure trig integral.

Steps

Step 1: Reflect $x\to a-x$.

If the denominator is reflection-invariant (e.g. $1+\sin(\pi-x)=1+\sin x$), then

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

Common Mistakes

Mistake 1: Reflecting the denominator wrongly.

Why it's wrong: $\sin(\pi-x)=\sin x$, so $1+\sin(\pi-x)=1+\sin x$ stays the same; assuming it changes breaks the cancellation of $x$. Correct approach: reflect, keep the denominator, and add to get $2I=\pi\int_0^\pi\frac{dx}{1+\sin x}$.

Mistake 2: Sign error at $x=\pi$ when evaluating $\tan x-\sec x$. …