Integration

Test Preparation
1

<p>Can someone explain to me in clear detail how to integrate the following:</p>

<p>y=54 + 23sin[2pi/52(x-12)]</p>

<p>-Thanks.</p>

2

<p>Hm. Okay, I'll try my best to help you with this problem but I have no guarantee that I will not make any mistakes since this is from my calculus AB summer self-study..Only been doing it for about two months. Here goes:</p>

<p>y=54 + 23sin[2pi/52(x-12)]</p>

<p>Integrate w.r.t. x:</p>

<p>∫ 54 + 23sin[2pi/52(x-12)] dx</p>

<p>Now, let's split this in two and integrate them separately. First tackle the 54:</p>

<p>54 is also equivalent to saying 54 * x^0
Therefore, using the basic integration law, we should add 1 to the power of the x and divide the entire thing by the final power:</p>

<p>∫ 54 * x^0 dx
= [(54 * x^1)/1] + c</p>

<h2>First part of the problem is done.</h2>

<p>Now, let's tackle the second portion:</p>

<p>∫ 23sin[2pi/52(x-12)] dx
= [23(-cos[2pi/52(x-12)]) * 1/ln(x-12)] + c</p>

<p>This is because sin becomes -cos when integrated.
Also, integration is the opposite of differentiation. The reason we have the 1/ln(x-12) is because when you differentiate this problem, you will have to pull out the ln(x-12). The 1/ln(x-12) is there to balance out the equation. (Sorry, I'm trying my best to explain it but it may come out as being quite confusing.)</p>

<p>--</p>

<p>Therefore, now we should combine both to give us:</p>

<p>∫ 54 + 23sin[2pi/52(x-12)] dx
= 54x - [23cos[2pi/52(x-12)] / ln(x-12)] + c</p>

<p>In the end, there is only 1 "c" since we can put the constants together. </p>

<p>If you have any points you need to clarify feel free to ask and I'll try my best to answer. I hope I did the question right. XD</p>

<p>--</p>

<p>Also, the way to test if the answer is correct is to differentiate it.</p>

3

<p>is it y=54+23sin[(2pi/52)(x-12)] or
y=54+23sin[2pi/(52(x-12))] ?</p>

<p>if it is the second one it does not look integrable</p>

<p>either way, I don't think you did it right</p>

4

<p>^Geez, mind explaining why instead of leaving the generous poster with doubts about his method?</p>

5

<p>It's the first one</p>

6

<p>I think Quesce solved the problem the second way Nalcon wrote it (with the x-12 in the denominator of the parameter of the sine function). When you solve it the second way here's what you get: </p>

<p>Do the first part of the problem like Quesce said. For the second part:
∫23sin[(2pi/52)(x-12)]<br>
= 23 * ∫sin[(2pi/52)(x-12)]
= 23 * <a href="52/2pi">color=#008800</a>[/color]* ∫[<a href="2pi/52">color=blue</a>[/color] * sin[(2pi/52)(x-12)]]
--------------<a href="multiply%20by%20%5Bcolor=blue%5D(2pi/52)%5B/color%5D%20so%20that%20you%20have%20∫%5Bcolor=red%5Df(u)%5B/color%5D%5Bcolor=blue%5Ddu%5B/color%5D.%20Multiply%20by%20the%20inverse%20%5Bcolor=#008800%5D(52/2pi)%5B/color%5D%20so%20the%20net%20effect%20is%20none.">/color</a>
= 23 * <a href="52/2pi">color=#008800</a> * -cos[(2pi/52)(x-12)]
= -(598/pi)*cos[(2pi/52)(x-12)]</p>

<p>So I think the final solution is F=54x-(598/pi)*cos[(2pi/52)(x-12)]</p>

7

<p>don't forget the +c</p>

8

<p>Ah okay. Learned from a mistake today. Thanks tanman :)</p>

<p>XD Sorry if I confused you YoungBuddhist.</p>