Animated Solution for Mathematics - Definite Integration: $6{\int }_0^{\pi }|(\sin 3x+\sin 2x+\sin x)|dx$ is equal to .........

A Journey Through the Modulus Integral

Imagine you are standing at the edge of a vast mathematical landscape, looking at the expression $6{\int }_0^{\pi }|\sin 3x+\sin 2x+\sin x|dx$. It looks intimidating, but in the world of JEE Advanced, complexity is often just a mask for elegance.

Our mission today is to peel back that mask.

The Anatomy of the Integrand

We begin by focusing on the function inside the modulus: $f(x)=\sin 3x+\sin 2x+\sin x$. Dealing with three separate sine terms is messy, so we must simplify.

We invoke the sum-to-product identity: $\sin C+\sin D=2\sin \frac{C+D}{2}\cos \frac{C-D}{2}$. Applying this to the first and third terms, we get:

Now, our function looks like this: $f(x)=2\sin 2x\cos x+\sin 2x$. Factoring out the common term $\sin 2x$, we obtain:

Using the double angle identity $\sin 2x=2\sin x\cos x$, we arrive at the fully factorized form:

Finding the Roots and the Sign

Now that we have the factored form, finding the roots is trivial. We set $f(x)=0$:

1. $\sin x=0\Rightarrow x=0,\pi$ 2. $\cos x=0\Rightarrow x=\frac{\pi }{2}$ 3. $2\cos x+1=0\Rightarrow \cos x=-\frac{1}{2}\Rightarrow x=\frac{2\pi }{3}$

These four points—$0,\frac{\pi }{2},\frac{2\pi }{3},\pi$—are the boundaries where our function changes its sign.

Between $0$ and $\frac{\pi }{2}$, all factors are positive, so $f(x)>0$. Between $\frac{\pi }{2}$ and $\frac{2\pi }{3}$, the $\cos x$ term turns negative, making $f(x)<0$. Finally, between $\frac{2\pi }{3}$ and $\pi$, both $\cos x$ and $(2\cos x+1)$ are negative, making $f(x)>0$ again.

The Integration

With the sign intervals established, we split the integral into three parts:

We calculate the general antiderivative $F(x)$:

Evaluating this at our boundaries: $F(0)=-\frac{11}{6}$, $F(\frac{\pi }{2})=\frac{1}{2}$, $F(\frac{2\pi }{3})=\frac{5}{12}$, and $F(\pi )=\frac{5}{6}$.

As we combine these values, the terms align and the fractions simplify. We find that the integral evaluates to $\frac{17}{6}$.

Finally, we multiply by the coefficient $6$ from the original problem. The $6$s cancel out, leaving us with the elegant integer 17.