An integral from IIT JEE 1995 entrance exam

 Back in the 20th century, the IIT JEE entrance exam used to have some pretty good subjective questions.

 Unfortunately, the format now is objective questions only, and these days it feels like the students in the coaching classes are taught to game the exams rather than learn the concepts: a regrettable outcome of removing the subjective questions. While one could argue that there are issues with having subjective questions, has it really been worthwhile?

Anyway, this is a cute integral problem from the 1995 JEE maths entrance exam.

Prove, using induction, that

$$\int_{0}^{\pi} \dfrac{1 - \cos mx} {1 - \cos x} \text {  dx} = m \pi$$

Scroll down for solutions.

.

.

.

.

.

1) $\textbf{A fully elementary induction based solution:}$ Everything in this solution falls in the syllabus for IIT JEE and we could expect students to be able to come up with this during the exam.

Let $$I_m = \int_{0}^{\pi} \dfrac{1 - \cos mx} {1 - \cos x} \text {  dx}$$

It is easy to verify that $I_m = m \pi$ for $m = 0, 1$.

Now $$I_{m+1} - I_m = \int_{0}^{\pi} \dfrac{\cos mx - \cos (m+1)x}{1 - \cos x}$$

Now using the formula $\cos 2A - \cos 2B = 2 \sin(A+B)\sin(B-A)$ and $1 - \cos x = 2 \sin^2 \left(\frac{x}{2}\right)$we get

$$I_{m+1} - I_m = \int_{0}^{\pi} \dfrac{2\sin ((m + \frac{1}{2})x) \sin \frac{x}{2}}{2\sin^2 \left(\frac{x}{2}\right)} = \int_{0}^{\pi} \dfrac{\sin ((m + \frac{1}{2})x)}{\sin \frac{x}{2}}$$

[A more knowledgeable reader will recognize the Dirichlet Kernel and cut short the rest of the proof]

Making the substitution $x = 2u$ gives us

$$ I_{m+1} - I_m = 2 \int_{0}^{\frac{\pi}{2}} \dfrac{\sin ((2m+1)u)}{\sin u} \text{ du}$$

Let $$J_m = \int_{0}^{\frac{\pi}{2}} \dfrac{\sin ((2m+1)u)}{\sin u} \text{ du}$$

Note that $2J_0 = \pi$. 

Now we have that

$$J_{m} - J_{m-1}= \int_{0}^{\frac{\pi}{2}} \dfrac{\sin ((2m+1)u) - \sin ((2m-1)u)}{\sin u} \text{ du}$$

Now using $\sin(A+B) - \sin(A-B) = 2 \cos A \sin B$ gives us

$$J_{m} - J_{m-1} = \int_{0}^{\frac{\pi}{2}} \dfrac{2 \cos (2mu) \sin u}{\sin u} = \int_{0}^{\frac{\pi}{2}} 2 \cos (2mu) =   0$$

Thus $J_m = J_0$, and so $I_{m+1} - I_{m} = 2J_m = 2J_0 = \pi$ and so $I_{m} = m \pi$ by induction.

2) $\textbf{Another elementary proof:}$  Show that $I_{m+1} + I_{m-1} = 2I_m$.

3) $\textbf{Possible advanced proof:}$ I suspect there will a quick proof based on fourier series. If you can find it, please comment.