A difficult substitution & separable integral

In summary: 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.In summary, the first question asks for the area of a surface obtained by rotating the curve y = cos(x), while the second asks to solve: \frac{dy}{dt} = \frac{ty+3t}{t^2+1}\ y(2)=2.
  • #1
silicon_hobo
59
0

Homework Statement


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 [tex]y = cos(x), 0 \leq x \leq\ \frac{\pi}{3}[/tex]

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

Homework Equations



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.
 
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  • #2
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
Thanks for the reply.

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

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

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
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 [itex]\pi/3[/itex]. 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 [itex]\pi/3[/itex] as your limits of integration. That's incorrect. When x= 0, u= 0 so the lower limit is still 0 but when x= [itex]\pi/3[/itex], u= sin([itex]\pi/3[/itex])= [itex]\sqrt{3}/2[/itex].
 
  • #5
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
[tex]\int\sec^3 xdx=\mbox{the average of the derivative and integral of secant}[/tex]
 
  • #7
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
rocomath said:
[tex]\int\sec^3 xdx=\mbox{the average of the derivative and integral of secant}[/tex]

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
silicon_hobo said:
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.

No, because you didn't substitute anything. You used integration by parts. Everything is in terms of x, not u or v.
 
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  • #10
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
bump (save me from these numbers and symbols)
 

1. What is a difficult substitution in integrals?

A difficult substitution is a technique used in integrals to simplify the integrand or make it easier to integrate. It involves substituting a variable or expression in the integrand with a more suitable one, which makes the integration process easier.

2. How do you identify when to use a difficult substitution in an integral?

You can identify when to use a difficult substitution by looking for complicated expressions or expressions that involve multiple variables in the integrand. Additionally, if the integrand contains a radical, trigonometric functions, or both, it may require a difficult substitution to simplify the expression.

3. What is a separable integral?

A separable integral is an integration technique used for integrals where the integrand can be separated into two functions, each depending on a different variable. This allows the integral to be written as the product of two simpler integrals, making it easier to solve.

4. How do you solve a separable integral?

To solve a separable integral, you need to first identify if the integrand can be separated into two functions. Then, use the appropriate techniques to integrate each function separately. Finally, multiply the results together to get the solution to the original integral.

5. What are some common examples of difficult substitution and separable integrals?

Some common examples of difficult substitution integrals include those involving trigonometric functions, exponential functions, and functions involving radicals. Separable integrals can be found in various applications, such as in physics, chemistry, and engineering, where the variables can be separated into distinct parts.

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