- #1

- 33

- 0

So far, I have ∫2ydy=-∫sin(3x)/cos^2(3x)dx. Is that right? If so, how do I integrate sin(3x)/cos^2(3x)?

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 Ric-Veda
- Start date

- #1

- 33

- 0

So far, I have ∫2ydy=-∫sin(3x)/cos^2(3x)dx. Is that right? If so, how do I integrate sin(3x)/cos^2(3x)?

- #2

fresh_42

Mentor

- 15,120

- 12,823

Is this homework? And do you mean ##\sin(3x) + 2\cos(3x) \cdot y(x) \cdot y'(x) = 0\;##?

- #3

- 33

- 0

These are practice questions for elementary differential equations. Oh...I just realized that there was a typo.Is this homework? And do you mean ##\sin(3x) + 2\cos(3x) \cdot y(x) \cdot y'(x) = 0\;##?

It's: Sin(3x)dx+2yCos^2(3x)dy=0

- #4

- 33

- 0

So far, I have ∫2ydy=-∫sin(3x)/cos^2(3x)dx. Is that right? If so, how do I integrate -∫sin(3x)/cos^2(3x)?

- #5

fresh_42

Mentor

- 15,120

- 12,823

E.g. it includes a paragraph for your own efforts, which shortens the process a lot. (Normally we don't want to give away ready made solutions to students but try to teach them instead.)

Your typo makes the entire equation more difficult, for otherwise it would have been simply the tangent.

You can look up many integrals here: https://de.wikibooks.org/wiki/Forme...timmte_Integrale_trigonometrischer_Funktionen

It's the wrong language but there is little beside a list of formulas. You won't need the language.

##\int \frac{\sin (3x)}{\cos^2 (3x)} = \frac{1}{3 \cos(3x)}## but I haven't checked whether it's correct.

- #6

- 33

- 0

Thank you for the link. It translates automatically into any language so it's ok. I wanted to know the whole steps on how to solve the integral: sin(3x)/cos^2(3x) (I found the template, but don't know how to use it yet. I don't go on that much on these forums for help. It's my first time using this to ask for help)

E.g. it includes a paragraph for your own efforts, which shortens the process a lot. (Normally we don't want to give away ready made solutions to students but try to teach them instead.)

Your typo makes the entire equation more difficult, for otherwise it would have been simply the tangent.

You can look up many integrals here: https://de.wikibooks.org/wiki/Forme...timmte_Integrale_trigonometrischer_Funktionen

It's the wrong language but there is little beside a list of formulas. You won't need the language.

##\int \frac{\sin (3x)}{\cos^2 (3x)} = \frac{1}{3 \cos(3x)}## but I haven't checked whether it's correct.

- #7

fresh_42

Mentor

- 15,120

- 12,823

No problem. Knowing the answer one could simply differentiate. The other way might be using one of the forms ##\frac{\sin x}{\cos^2 x}=\frac{\tan x}{\cos x} ## and integrate by parts: ##\int u'v = uv - \int uv'##.Thank you for the link. It translates automatically into any language so it's ok. I wanted to know the whole steps on how to solve the integral: sin(3x)/cos^2(3x) (I found the template, but don't know how to use it yet. I don't go on that much on these forums for help. It's my first time using this to ask for help)

And thanks for the language test. I didn't know this, but surely good to know as - you might have recognized it - there are a lot more formulas for all kind of integrals.

- #8

- 22,129

- 3,297

Look up Weierstrass substitution.

- #9

- 33

- 0

Thanks, I know how to do it now.No problem. Knowing the answer one could simply differentiate. The other way might be using one of the forms ##\frac{\sin x}{\cos^2 x}=\frac{\tan x}{\cos x} ## and integrate by parts: ##\int u'v = uv - \int uv'##.

And thanks for the language test. I didn't know this, but surely good to know as - you might have recognized it - there are a lot more formulas for all kind of integrals.

Share: