Animated Solution for Mathematics - Definite Integration: ${\int }_0^{20\pi }(|\sin x|+|\cos x|{)}^{2}dx$ is equal to :-

Welcome, future IITian. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of absolute values and trigonometric functions.

You might look at the integral and feel a surge of anxiety. That is perfectly normal. The secret to JEE Advanced is not to fight the complexity head-on, but to simplify the battlefield until the solution reveals itself.

The first thing we notice is the square. We have $(|\sin x|+|\cos x|{)}^2$. In algebra, we know that $(a+b{)}^2=a^2+b^2+2ab$.

Applying this identity, we get:

Here is the first moment of clarity. Remember that for any real number $x$, $|x{|}^2=x^2$. The absolute value becomes irrelevant when you square the term.

So, our expression simplifies to:

Look at that! ${\sin }^{2}x+{\cos }^{2}x$ is the most famous identity in trigonometry, equal to $1$. We have already reduced the complexity significantly to $1+2|\sin x||\cos x|$.

We are not done yet. We have the term $2|\sin x||\cos x|$. We can combine the absolute values because $|a||b|=|ab|$.

This gives us $|2\sin x\cos x|$. Does that ring a bell? It should! The double-angle identity tells us that $2\sin x\cos x=\sin 2x$.

So, our entire integrand has collapsed from a terrifying expression into the elegant function:

Now, we look at the limits: $0$ to $20\pi$. Integrating this over such a large range would be a nightmare if we tried to do it manually. This is where we use the JEE educator's favorite tool: periodicity.

We know the period of $\sin x$ is $2\pi$. Therefore, the period of $\sin 2x$ is $\pi$. But wait, we have the absolute value! The absolute value of $\sin 2x$ flips the negative parts of the wave up, creating a new peak every $\pi /2$.

Thus, the period of our function $f(x)=1+|\sin 2x|$ is $T=\pi /2$. We need to find how many such periods fit into our interval of $20\pi$.

We calculate the number of cycles $n$:

The property of definite integrals for periodic functions states that:

This means our integral is simply:

We are in the home stretch. We need to integrate $1+|\sin 2x|$ from $0$ to $\pi /2$. In this interval, $2x$ ranges from $0$ to $\pi$.

In the first and second quadrants, $\sin 2x$ is always positive. This means we can safely remove the absolute value bars:

The integral of $1$ is $x$, and the integral of $\sin 2x$ is $-\frac{\cos 2x}{2}$. Evaluating this from $0$ to $\pi /2$, we get:

Substituting the upper limit:

Substituting the lower limit:

Subtracting the lower from the upper, we get:

Finally, multiply by our $40$ cycles:

Factoring out the $20$, we arrive at the final answer: $20(\pi +2)$

You did it. You took a complex problem, broke it down, applied the right identities, used the power of periodicity, and arrived at the solution. This is the mindset of a topper. Keep practicing, and keep falling in love with the process.