- #1

- 4

- 0

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!