# Homework Help: A difficult substitution & separable integral

1. Apr 1, 2008

### silicon_hobo

1. The problem statement, all variables and given/known data
I'm not sure how to proceed here. The first one asks me to find the area of a surface obtained by rotating the curve $$y = cos(x), 0 \leq x \leq\ \frac{\pi}{3}$$

The second one asks to Solve: $$\frac{dy}{dt} = \frac{ty+3t}{t^2+1}\ y(2)=2$$

2. Relevant equations

3. The attempt at a solution
http://www.mcp-server.com/~lush/shillmud/int3.4.JPG
I've tried a bunch of different substitutions but I always stall on the integration.

http://www.mcp-server.com/~lush/shillmud/int3.7.JPG
I'm not sure how to satisfy the condition y(2) = 2. Thank you again for your help.

Last edited: Apr 2, 2008
2. Apr 1, 2008

### Snazzy

For the first question, integrate (secx)^3 by parts with u = secx and dv = (secx)^2.

For the second question, remember, there's a + C.

3. Apr 2, 2008

### silicon_hobo

Alright so if I inetgrate the first one fully from sec^3 I get:

$$\int^\frac{\pi}{3}_{0}(sec^3\ x\ dx) = sec\ x\ tan x - \int(sec\ x\ tan^2\ x dx)$$
$$= sec\ x\ tan\ x - \int(sec\ x\ (sec^2\ x - 1) dx)$$
$$= sec\ x\ tan\ x - \int(sec^3\ x\ dx) + \int(sec\ x\ dx)$$
$$= sec\ x\ tan\ x - \int(sec^3\ x\ dx) + ln(sec\ x + tan\ x) + C$$
$$\int(sec^3\ x\ dx) = 1/2(sec\ x\ tan\ x + ln(sec\ x + tan\ x)) + C$$

But as you can see in my orginal post I have been a substitution. I believe I must return to the original term. However, I do not see in opportunity to do so during the integration above.

As for the second question, how does C change anything there? Cheers.

4. Apr 3, 2008

### HallsofIvy

Why did you change from a definite integral to an indefinite integral?

The way you have this written, yes, you should go back to your original variable, x, and evaluate at 0 and $\pi/3$. The "C" will cancel and means nothing in a definite integral.

It's usually simpler to change the limits of integration as you substitute.

In what you wrote originally you make the substution u= -sin(x) but still has 0 and $\pi/3$ as your limits of integration. That's incorrect. When x= 0, u= 0 so the lower limit is still 0 but when x= $\pi/3$, u= sin($\pi/3$)= $\sqrt{3}/2$.

5. Apr 4, 2008

### silicon_hobo

Right, but do I have to change the limits of integration again for the 2nd sub to tan & sec^2... seems like I have done this one the hard way.

6. Apr 4, 2008

### rocomath

$$\int\sec^3 xdx=\mbox{the average of the derivative and integral of secant}$$

7. Apr 4, 2008

### rohanprabhu

i have a doubt here. When we say that a curve is rotated, like on rotation of y = cos(x), as in this question.. which axis are we rotating it along?

I could rotate it along the x-axis, the y-axis or an axis that is perpendicular to the y-axis and passes through (0, Cos[pi/3]).

8. Apr 4, 2008

### silicon_hobo

Ok, so is that a yes or no to the "do I need to change the limits of integration a second time" question:P Thanks for your patience folks. It's a rotation about the x-axis.

9. Apr 4, 2008

### Snazzy

No, because you didn't substitute anything. You used integration by parts. Everything is in terms of x, not u or v.

Last edited: Apr 4, 2008
10. Apr 4, 2008

### silicon_hobo

Sorry, I've made this really confusing (for myself as well). When i put the integration of sec^3 in latex i used x but if you look to the top it should actually be an integration of sec^3 theta which is the result of a substitution of -sin x for tan theta.

11. Apr 5, 2008

### silicon_hobo

bump (save me from these numbers and symbols)