- #1

CellCoree

- 42

- 0

i just need to know how to start this, i can't use u-du...

i can't borrow anything...

so how would i start this?

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 CellCoree
- Start date

- #1

CellCoree

- 42

- 0

i just need to know how to start this, i can't use u-du...

i can't borrow anything...

so how would i start this?

- #2

Tide

Science Advisor

Homework Helper

- 3,089

- 0

Is the integrand the same as

[tex]\frac {\cos {6x} - \cos {20x}}{2}[/tex]

?

[tex]\frac {\cos {6x} - \cos {20x}}{2}[/tex]

?

- #3

teknodude

- 157

- 0

- #4

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,967

- 23

Another approach, if you've learned the necessities, is to replace the sine function with its definition in terms of complex exponentials.

Another way is to use the angle sum identities. Your problem is that 7x and 13x have different coefficients, so break 13x into 7x + 6x and use the sum of angles identity for sine. Rinse, and repeat until you can do all of the resulting integrals. One thing you should

- #5

ehild

Homework Helper

- 15,543

- 1,915

Tide said:Is the integrand the same as

[tex]\frac {\cos {6x} - \cos {20x}}{2}[/tex]

?

YES!

You know, [tex]cos(\alpha +\beta )= cos(\alpha )cos(\beta ) - sin(\alpha )sin(\beta) \mbox { and } cos(\alpha -\beta )= cos(\alpha )cos(\beta ) + sin(\alpha )sin(\beta)\mbox. \\ [/tex]

[tex]\alpha = 13x \mbox{ and } \beta = 7x [/tex]...

ehild

Share:

- Last Post

- Replies
- 2

- Views
- 762

- Replies
- 5

- Views
- 321

- Replies
- 5

- Views
- 565

- Replies
- 19

- Views
- 543

- Last Post

- Replies
- 4

- Views
- 236

- Last Post

- Replies
- 9

- Views
- 448

- Last Post

- Replies
- 10

- Views
- 366

- Replies
- 5

- Views
- 379

- Replies
- 7

- Views
- 402

- Replies
- 2

- Views
- 1K