Evaluate $\int_0^{\pi/2} \cos^2(wt)dt$ where w is constant

Question

We can proceed by $$\frac{\int_0^{\pi/2}(\cos2wt+1)dt}{2}$$

(I hope you don’t mind if don’t write the limits for the next few steps, as it is becoming cumbersome)

So, $$\frac12\int \cos2wt+\frac12\int 1$$ $$\frac{1}{4w}\sin2wt+\frac t2$$ Putting in the limits, we end up with $$\frac12\Bigl[\frac{1}{2w}+\frac{\pi}{2}\Bigr]$$

That’s as far as I could go. The answer is $\frac{\pi}{4}$. So either I have gone wrong in one step or there is a way to proceed further. Thanks for helping!

Yes, you are right. My bad. Although my problem still stands — Aditya, Aug 13 '19 at 15:41
The answer you expect is valid only if $w$ is an integer. I don't understand why you choose $\sin 2w\frac{\pi}{2}$ to be $1$. If $w\in\mathbb Z$ then $\sin w\pi=0$ — Andrei, Aug 13 '19 at 15:51

score 1 · Answer 1 · answered Aug 13 '19 at 15:37

1

Hint: $$\int\cos(2wt)dt=\frac{\sin(2wt)}{2w}+C$$

answered Aug 13 '19 at 15:37

Dr. Sonnhard Graubner

95,283

But it should also work with the formula I used – Aditya Aug 13 '19 at 15:40

score 1 · Accepted Answer · answered Aug 13 '19 at 16:17

The integration step was correct now let $J$ the remaining, we have: $J=[\frac{1}{4w}\sin(2wt) + \frac{t}{2}]_{0}^{\pi/2} =\frac{1}{4w}\sin(w\pi)+\pi/4 -0=\frac{\sin(w\pi)}{4w}+\frac{\pi}{4}$ If $w\in\mathbb{Z} \Rightarrow \sin(w\pi)=0$ Therefore $J=0+\pi/4=\pi/4$

Evaluate $\int_0^{\pi/2} \cos^2(wt)dt$ where w is constant

2 Answers2