Animated Solution for Mathematics - Definite Integration: Let $I_m={\int }_0^{\pi }\frac{1-\cos mx}{1-\cos x}dx$. Use mathematical induction to prove that $I_m=m\pi ,m=0,1,2,\dots$.

Welcome, future engineers! Today, we are not just solving an integral; we are embarking on a journey through the elegance of mathematical induction.

We are tasked with proving that the integral

is equal to $m\pi$ for all non-negative integers $m$. This problem is a classic, and it hides a beautiful symmetry that reveals itself only when we approach it with patience and the right tools.

Every great structure needs a solid foundation. In induction, this is our base case.

We start by testing $m=0$. When we substitute $m=0$ into our integral, the numerator becomes $1-\cos (0)$, which is $1-1=0$. The integral of zero is simply zero, and our formula $m\pi$ also gives $0\cdot \pi =0$.

Because our recurrence relation will eventually link three consecutive terms ($I_{k+1},I_k,I_{k-1}$), we need a second base case to get the engine running. Let us test $m=1$:

Since our formula $m\pi$ gives $1\cdot \pi =\pi$, both base cases are rock solid.

Now, we assume the formula holds for two consecutive integers, $k-1$ and $k$. That is, we assume $I_{k-1}=(k-1)\pi$ and $I_k=k\pi$.

Our mission is to prove that $I_{k+1}=(k+1)\pi$. To bridge this gap, we look at the combination $I_{k+1}+I_{k-1}-2I_k$.

When we write this as a single integral, the numerator becomes:

The constant terms $1+1-2$ vanish, leaving us with $2\cos kx-\cos (k+1)x-\cos (k-1)x$.

This is where the beauty of trigonometry shines. We have the term $\cos (k+1)x+\cos (k-1)x$.

Using the sum-to-product identity $\cos (A+B)+\cos (A-B)=2\cos A\cos B$, we can transform this into $2\cos kx\cos x$. Now, look at our numerator:

When we place this back over our denominator, $1-\cos x$, the entire fraction simplifies beautifully:

Since ${\int }_0^{\pi }\cos kxdx=0$ for any integer $k\ge 1$, we have $I_{k+1}+I_{k-1}-2I_k=0$, or $I_{k+1}=2I_k-I_{k-1}$.

With our recurrence relation $I_{k+1}=2I_k-I_{k-1}$ in hand, we substitute our inductive hypotheses:

Simplifying this, we get $2k\pi -k\pi +\pi$, which is exactly $(k+1)\pi$.

We have proven that if the formula works for two steps, it must work for the next. By the principle of mathematical induction, the formula $I_m=m\pi$ is true for all non-negative integers $m$.