- #1
Captain Zappo
- 17
- 0
Hi everyone. I'm having some trouble evaluating the following integral
[tex]
\int{sin^4xdx}
[/tex]
First let me start off by showing what I did.
[tex]
= \int{(sin^2x)(sin^2x)dx}
[/tex]
[tex]
=\int{[\frac{1}{2}-\frac{1}{2}cos(2x)] \ [\frac{1}{2}-\frac{1}{2}cos(2x)]dx
[/tex]
[tex]
=\int{(\frac{1}{4}-cos(2x)+\frac{1}{4}cos^2(2x))dx
[/tex]
[tex]
=\frac{1}{4}x-\frac{1}{2}sin(2x)+\frac{1}{4}\int{(\frac{1}{2}+\frac{1}{2}cos(4x))dx
[/tex]
[tex]
=\frac{1}{4}x-\frac{1}{2}sin(2x)+\frac{1}{8}x+\frac{1}{8}\int{cos(4x)}dx
[/tex]
[tex]
=\frac{1}{4}x-\frac{1}{2}sin(2x)+\frac{1}{8}x+\frac{1}{32}sin(4x)+C
[/tex]
I'm not sure if I have the right answer or not, so could someone please check and see if I did anything illegal. If so, can you please correct me, or even show me a different method entirely.
Thanks,
-Zach
[tex]
\int{sin^4xdx}
[/tex]
First let me start off by showing what I did.
[tex]
= \int{(sin^2x)(sin^2x)dx}
[/tex]
[tex]
=\int{[\frac{1}{2}-\frac{1}{2}cos(2x)] \ [\frac{1}{2}-\frac{1}{2}cos(2x)]dx
[/tex]
[tex]
=\int{(\frac{1}{4}-cos(2x)+\frac{1}{4}cos^2(2x))dx
[/tex]
[tex]
=\frac{1}{4}x-\frac{1}{2}sin(2x)+\frac{1}{4}\int{(\frac{1}{2}+\frac{1}{2}cos(4x))dx
[/tex]
[tex]
=\frac{1}{4}x-\frac{1}{2}sin(2x)+\frac{1}{8}x+\frac{1}{8}\int{cos(4x)}dx
[/tex]
[tex]
=\frac{1}{4}x-\frac{1}{2}sin(2x)+\frac{1}{8}x+\frac{1}{32}sin(4x)+C
[/tex]
I'm not sure if I have the right answer or not, so could someone please check and see if I did anything illegal. If so, can you please correct me, or even show me a different method entirely.
Thanks,
-Zach
Last edited: