- #1

- 382

- 0

What is the best way?

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- Thread starter kasse
- Start date

- #1

- 382

- 0

What is the best way?

- #2

- 1,753

- 1

Idk, you didn't show any work.

- #3

- 382

- 0

(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

- 1,753

- 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

- 382

- 0

not cosine, sine

- #6

- 1,753

- 1

What are you talking about?not cosine, sine

- #7

- 382

- 0

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

- #8

- 1,753

- 1

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

- #9

- 382

- 0

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

- 1,753

- 1

distribute the sinx and you'll see your solution

- #11

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 966

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"!

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