- #1

kasse

- 382

- 1

What is the best way?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter kasse
- Start date

- #1

kasse

- 382

- 1

What is the best way?

- #2

rocomath

- 1,755

- 1

Idk, you didn't show any work.

- #3

kasse

- 382

- 1

(1-cos^2(x))*sin(x)

u = 1- cos^2(x)

du/dx = -2cos(x)sin(x)

so that

(1-cos^2(x))*sin(x) dx = (-u sin(x)/2cos(x)) du

Doesn't really help, or?

- #4

rocomath

- 1,755

- 1

[tex]\int\cos x\cos^2 xdx[/tex]

Use a BASIC trig identity to change the 2nd degree cosine function.

Use a BASIC trig identity to change the 2nd degree cosine function.

- #5

kasse

- 382

- 1

not cosine, sine

- #6

rocomath

- 1,755

- 1

What are you talking about?not cosine, sine

- #7

kasse

- 382

- 1

Don't know, I guess I'm too drunk to do maths right now.

- #8

rocomath

- 1,755

- 1

Try again later :)Don't know, I guess I'm too drunk to do maths right now.

- #9

kasse

- 382

- 1

sin^3 x

=

sin^2 x*sin x

=

(1 - cos^2 x)sin x

Then substitution?

u = 1-cos^2 x

du/dx = 2cos x*sin*x

so that

sin^3 x dx = - u / 2cos x

Hm...

- #10

rocomath

- 1,755

- 1

distribute the sinx and you'll see your solution

- #11

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 970

So how about just u= cos(x)?

sin^3 x

=

sin^2 x*sin x

=

(1 - cos^2 x)sin x

Then substitution?

u = 1-cos^2 x

du/dx = 2cos x*sin*x

so that

sin^3 x dx = - u / 2cos x

Hm...

You know what they say "Don't drink and derive"!

Share:

- Last Post

- Replies
- 5

- Views
- 262

- Replies
- 44

- Views
- 2K

- Last Post

- Replies
- 6

- Views
- 415

- Last Post

- Replies
- 9

- Views
- 156

- Replies
- 29

- Views
- 746

- Replies
- 18

- Views
- 1K

- Replies
- 2

- Views
- 394

- Replies
- 7

- Views
- 619

- Replies
- 1

- Views
- 541

- Replies
- 6

- Views
- 685