Animated Solution for Mathematics - Definite Integration: The value of integral ${\int }_{\pi /4}^{3\pi /4}\frac{x}{1+\sin x}dx$ is :-

Taming the Troublemaker

Imagine you are standing before a complex integral:

At first glance, the $x$ in the numerator feels like a persistent troublemaker. It prevents us from using simple substitution or standard trigonometric identities.

But in the world of JEE Advanced, every 'troublemaker' has a weakness. Our goal today is to find that weakness and dismantle this problem with elegance.

A Strategic Shift

Whenever you see a definite integral where an $x$ is causing trouble, your first instinct should be the King's Property:

This property is not just a formula; it is a geometric insight. It tells us that the area under the curve $f(x)$ from $a$ to $b$ is the same as the area under the curve $f(a+b-x)$ over the same interval.

Here, our limits are $a=\pi /4$ and $b=3\pi /4$. Their sum is $a+b=\pi$. So, we replace every $x$ with $(\pi -x)$.

Let us perform the transformation:

Because $\sin (\pi -x)=\sin x$, our integral becomes:

Now, watch the magic. If we add our original integral $I$ to this new version, we get:

The $x$ and $-x$ cancel out perfectly! We are left with:

The troublemaker is gone.

Now we face the integral of $\frac{1}{1+\sin x}$. The standard trick here is rationalization.

Multiply the numerator and denominator by $(1-\sin x)$ to get:

Since $1-{\sin }^{2}x={\cos }^{2}x$, this simplifies to:

These are standard integrals! The integral of ${\sec }^{2}x$ is $\tan x$, and the integral of $\sec x\tan x$ is $\sec x$. Thus:

We must be precise with our limits. At the upper limit $x=3\pi /4$:

At the lower limit $x=\pi /4$:

Subtracting these, we get:

Finally, $2I=\pi \cdot 2(\sqrt{2}-1)$, which simplifies to:

$$I=\pi (\sqrt{2}-1)$$

We have successfully navigated the complexity and arrived at the elegant solution. Keep practicing these transformations; they are the key to mastering calculus.