Tricky trig integral (calc 1/2 level)

  • #1
Hello,

I am an incoming EE freshman who completed Calculus BC one year ago and Multivariable Calculus last Fall. Right now I'm doing a self-study going over basic calculus again in one of my review books.

Here is my problem:
[tex]\int sin^3(x)cos^5(x) dx[/tex]

and I solved it, with an answer my graphing calculator showed to be non-equivalent to the book answer:

= [tex]\int sin(x)(\frac{1-cos(2x)}{2})(\frac{1+cos(2x)}{2})^2cos(x) dx[/tex]

(expanding terms) = [tex]\int sin(x)cos(x)(\frac{1+cos(2x)-cos^2(2x)-cos^3(2x)}{8}) dx[/tex]

using sin(2x) = 2sin(x)cos(x), I integrated:

[tex]\frac{1}{8} \int \frac{sin(2x)}{2} + \frac{sin(2x)}{2}cos(2x) - \frac{sin(2x)}{2}cos^2(2x) - \frac{sin(2x)}{2}cos^3(2x) dx [/tex]

u = 2x, du = 2dx
u = sin(2x), du = 2cos(2x)dx
u = cos(2x), du = -2sin(2x)dx

My solution = [tex](1/16)sin^2(2x) - (1/32)cos(2x) + (1/24)cos^3(2x) + (1/32)cos^4(2x) + C[/tex]

Their solution = [tex]-(1/6)cos^6(x) + (1/8)cos^8(x) + C[/tex]

Like I said I graphed the solutions on my TI89 and they are not equivalent. What is also very difficult for me to understand is that I graphed my expansion of the integral vs the original integral, showing them to be equal. I then used the integration function to graph the integrals of the original and my expansion and they were shown to be different!

If there is an easier way to do that problem I would love to see it but I would also like to understand what if anything is wrong with my approach. Thank you!
 

Answers and Replies

  • #2
dextercioby
Science Advisor
Homework Helper
Insights Author
13,104
652
[tex] \int \sin^3 x \cos^5 x {}dx =\int \sin^3 x \cos^4 x {}\cos x {} dx =\int \sin^3 x \left(1-\sin^2 x\right)^2 d\sin x [/tex]

can you carry on from here ?
 
  • #3
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
10,025
134
What do you mean by "not equivalent"?

DO remember that valid antiderivatives can differ by a constant.

So, if your anti-derivative is a parallell curve to the one TI-89 spat out, you have done it right after all.
 
  • #4
dexter: I dont know why that didn't come to mind it is much simpler! (of course though you mean the last term to be cosx). thank you

and arildno: I had just looked at the table of values instead of the graph, and actually both solutions have a constant difference of 1/48 so it is actually correct.

thanks!! problem solved
 
  • #5
dextercioby
Science Advisor
Homework Helper
Insights Author
13,104
652
The last term is the differential one form [itex] d\sin x [/itex] which is nothing but the element of intergration.
 
  • #6
Gib Z
Homework Helper
3,346
6
Alternatively, [tex]\int \sin^3 x \cos^5 x dx = \int \sin x \sin^2 x \cos^5 x dx = \int (\cos^5 x - \cos^7 x) \sin x dx[/tex]
 
  • #7
The last term is the differential one form [itex] d\sin x [/itex] which is nothing but the element of intergration.

I am new to that notation. I suppose that would be the same as leaving dsinx as cosx and integrating with u-substitution?
 
  • #8
HallsofIvy
Science Advisor
Homework Helper
41,847
964
I am new to that notation. I suppose that would be the same as leaving dsinx as cosx and integrating with u-substitution?

It would be the same as d(sin x)= cos x dx.
 

Related Threads on Tricky trig integral (calc 1/2 level)

  • Last Post
Replies
6
Views
2K
H
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
675
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
1
Views
1K
Top