- #1
mbrmbrg
- 496
- 2
(1)[tex]\int\sin^3{x}\cos^3{x}dx[/tex]
(2)[tex]=\int\sin^3{x}(1-\sin^2{x})\cos{x}dx[/tex]
(3a) let u=sin(x), du=cos(x)dx
(3b)[tex]=\int(u^3-u^5)du[/tex]
(4)[tex]=\frac{u^4}{4} -\frac{u^6}{6} + C[/tex]
(5)[tex]=\frac{1}{4}\sin^4{x} -\frac{1}{6}\sin^6{x} + C[/tex]
If at step (2) I pull off a sin(x) and go from there, I end up with [tex]\frac{1}{6}\cos^6{x} -\frac{1}{4}\cos^4{x} + C[/tex]
I tried to prove that this is equvalent to step (5) by using sin^2(A)+cos^2(A)=1, but this turned step (5) into [tex]\frac{1}{12} +\frac{3}{4}\cos^4{x} -\frac{1}{6}\cos^6{x} + C[/tex]. And so I was bewildered and befuddled.
Then I decided to plug the original problem into a website's Integrator. And the integrator told me that the answer really is [tex]\frac{1}{192}(-9\cos(2x) +\cos(6x))[/tex].
Now I am just
Help, please, and be blessed!
(Last little bit of frustration: I typed up this post before I logged in, and even the browser's back button couldn't help me retrieve it. Latex is cool but twice?)
(2)[tex]=\int\sin^3{x}(1-\sin^2{x})\cos{x}dx[/tex]
(3a) let u=sin(x), du=cos(x)dx
(3b)[tex]=\int(u^3-u^5)du[/tex]
(4)[tex]=\frac{u^4}{4} -\frac{u^6}{6} + C[/tex]
(5)[tex]=\frac{1}{4}\sin^4{x} -\frac{1}{6}\sin^6{x} + C[/tex]
If at step (2) I pull off a sin(x) and go from there, I end up with [tex]\frac{1}{6}\cos^6{x} -\frac{1}{4}\cos^4{x} + C[/tex]
I tried to prove that this is equvalent to step (5) by using sin^2(A)+cos^2(A)=1, but this turned step (5) into [tex]\frac{1}{12} +\frac{3}{4}\cos^4{x} -\frac{1}{6}\cos^6{x} + C[/tex]. And so I was bewildered and befuddled.
Then I decided to plug the original problem into a website's Integrator. And the integrator told me that the answer really is [tex]\frac{1}{192}(-9\cos(2x) +\cos(6x))[/tex].
Now I am just
Help, please, and be blessed!
(Last little bit of frustration: I typed up this post before I logged in, and even the browser's back button couldn't help me retrieve it. Latex is cool but twice?)