Trigonometric Integration -split https://www.freemathhelp.com/forum/threads/106585-Ex

[ray5450]

Okay, thanks. I see the 2 forms are the same and correct.


Are these 2 forms correct:

⌡cos²x · dx = - cos³x + C
......................3sinx

or

................= sin2x + x + C
.................... 4........2

(Identity: cos²x = cos2x + 1)
................................2

 

[Deleted member 4993]

Okay, thanks. I see the 2 forms are the same and correct.


Are these 2 forms correct:

⌡cos²x · dx = - cos³x + C ...........→ Incorrect
......................3sinx

or

................= sin2x + x + C...........→ Correct
.................... 4........2

(Identity: cos²x = cos2x + 1)
................................2

.

 

[ray5450]

Is this correct:

y=cos³x

dy = 3cos²x (-sinx)
dx

 

[Harry_the_cat]

Is this correct:

y=cos³x

dy = 3cos²x (-sinx)
dx

Yes it is.

 

[Deleted member 4993]

Is this correct:

y=cos³x

dy = 3cos²x (-sinx)
dx

As Cat indicated that above statement correct - but this will not provide correct path for integration in your original post.

 

[ray5450]

If y=cos³x + C is the reverse of dy = 3cos²x (-sinx), then why can y= -cos³x + C not be a reverse of dy = cos²x ?
...........................................dx..................................................3sinx ..................................dx

 

[Deleted member 4993]

If y=cos³x + C is the reverse of dy = 3cos²x (-sinx), then why can y= -cos³x + C not be a reverse of dy = cos²x ?
...........................................dx..................................................3sinx ..................................dx

If you differentiate:

y = \(\displaystyle \dfrac{cos^3(x)}{sin(x)}\)

when:

y = \(\displaystyle \dfrac{f(x)}{g{x}}\)

\(\displaystyle \dfrac{dy}{dx} \ = \dfrac{f'*g - f*g'}{g^2}\)

In your case, f(x) = cos^3(x) and g(x) = sin(x)

\(\displaystyle \dfrac{dy}{dx} = \dfrac{cos^3(x)}{sin(x)} = \dfrac{3*cos^2(x) * [-sin^2(x)] - cos^3(x) * cos(x)}{sin^2(x)}\)

Does not look anything like cos^2(x) - does it?

 

[ray5450]

Ah... Apparently, there must me some rule allowing only constants in the denominator when integrating. My reference materials didn't mention that.

Thank-you.