Animated Solution for Mathematics - Definite Integration: If $g(x)={\int }_0^x{\cos }^{4}tdt$, then $g(x+\pi )$ equals

Unlocking $g(x+\pi )$

Welcome, future engineer! Today, we are going to look at a problem that might seem like a standard calculus exercise, but it is actually a masterclass in understanding how functions behave. We are given $g(x)={\int }_0^x{\cos }^{4}tdt$.

Our goal is to find $g(x+\pi )$. Don't reach for your pen to start integrating ${\cos }^{4}t$ using reduction formulas. If you do that, you are walking into a trap. In JEE Advanced, the most elegant path is rarely the one that requires the most algebra.

Visualizing the Area

Think of $g(x)$ as a 'bucket' that collects area. As $x$ increases, the bucket fills up with the area under the curve $y={\cos }^{4}t$. When we ask for $g(x+\pi )$, we are simply asking: 'What is the total area collected from $t=0$ all the way to $t=x+\pi$?'

We can use the additive property of definite integrals to break this journey into two distinct legs. We can stop at $\pi$ and then continue from $\pi$ to $x+\pi$. Mathematically, we write this as:

Look at the first term: ${\int }_0^{\pi }{\cos }^{4}tdt$. By the very definition of our function $g(x)$, this is just $g(\pi )$. We have already simplified half the problem!

The Art of Substitution

Now, let's focus on the second integral: $I={\int }_{\pi }^{x+\pi }{\cos }^{4}tdt$. This looks slightly intimidating because the limits are shifted. But remember, in calculus, we love symmetry.

Let's perform a substitution to shift these limits back to the origin. Let $t=y+\pi$. When $t=\pi$, $y=0$. When $t=x+\pi$, $y=x$. Our differential $dt$ becomes $dy$. The integral transforms into:

The Symmetry Triumph

Here is where the magic happens. We know from trigonometry that $\cos (y+\pi )=-\cos y$. But look at the power! We are dealing with ${\cos }^4(y+\pi )$.

This means we are calculating $(-\cos y{)}^4$. Because the exponent is even, the negative sign vanishes into thin air! We are left with ${\cos }^{4}y$. Our integral $I$ becomes:

Since $y$ is just a dummy variable, this is identical to ${\int }_0^x{\cos }^{4}tdt$, which is exactly $g(x)$.

We have arrived! We split the integral into $g(\pi )$ and the second part, which we proved is $g(x)$. Putting it all together, we get:

It is elegant, it is clean, and it shows that the area under this curve grows in a perfectly predictable, additive way. You didn't need to perform a single complex integration. You just needed to understand the geometry of the function. Keep this mindset—always look for the structure before you start the calculation!