Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm stuck on integration of the following function:

$$ \int_0^{2\pi} \frac{5}{2}|\sin(2t)| dt$$ I understand the few first steps with substitution, etc., but I can't get the end result which is "10".

Could someone give me a step by step solution for it?

share|improve this question
    
Draw the picture. We want $4$ times area from $t=0$ to $t=\pi/2$, or alternately $8$ times area from $0$ to $\pi/4$. –  André Nicolas May 20 '12 at 20:51

2 Answers 2

up vote 5 down vote accepted

Note that $$\lvert \sin(2t) \rvert = \begin{cases} \sin(2t) & t\in[0,\pi/2]\\ - \sin(2t) & t\in[\pi/2,\pi]\\ \sin(2t) & t\in[\pi,3\pi/2]\\ -\sin(2t) & t\in[3\pi/2, 2\pi] \end{cases}$$ Also, $$\int_0^{\pi/2} \sin(2t) dt = -\int_{\pi/2}^{\pi} \sin(2t) dt = \int_{\pi}^{3\pi/2} \sin(2t) dt = -\int_{3\pi/2}^{2\pi} \sin(2t) dt$$ Hence, the integral $$I = \int_0^{2 \pi} \frac52 \lvert \sin(2t) \rvert dt = 4 \times \int_0^{\pi/2} \frac52 \sin(2t) dt = 10 \times \int_0^{\pi/2} \sin(2t) dt = 10$$

share|improve this answer
    
FGITW - beat me to it. +1 –  mixedmath May 20 '12 at 18:48

We have $$\begin{equation} \int^{t=2 \pi}_{t=0} \frac{5}{2} |\text{sin} 2t | \, dt = \frac{5}{2} \int^{t=2 \pi}_{t=0} |\text{sin} 2t | \, dt \end{equation}$$ Now use the substitution $y = 2t$, $dy = 2dt$. The boundaries shift as follows: if $t = 2\pi$ then $y = 4\pi$ and if $t = 0$ then $y = 0$

We plug this into the integral to get $$\begin{equation} \frac{5}{2} \int^{t=2 \pi}_{t=0} |\text{sin} 2t | \, dt = \frac{5}{2} \int^{y=4 \pi}_{y=0} |\text{sin } y | \, \frac{dy}{2} = \frac{5}{4} \int^{y=4 \pi}_{y=0} |\text{sin } y | \, dy \end{equation}$$

It remains to remember that sin$x \geq 0$ for $x \in [0,\pi]$, and sin$x \leq 0$ when $ x \in [\pi, 2\pi]$. Finally, we can also use that sin$x$ is periodic, which means that sin$(x + 2\pi) = $ sin$x$.

So we get $$\begin{equation} \frac{5}{4} \int^{y=4 \pi}_{y=0} |\text{sin } y | \, dy = 2 \left(\frac{5}{4} \int^{y=2 \pi}_{y=0} |\text{sin } y | \, dy \right) = 2 \left(\frac{5}{4} \int^{y= \pi}_{y=0} \text{sin } y \, dy - \frac{5}{4} \int^{y= 2 \pi}_{y=\pi} \text{sin } y \, dy \right) \end{equation}$$

You can now compute the integral to obtain the desired result.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.