Finding the Indefinite Integral Extension Questions

[CaptainK]

Homework Statement

∫8x³e^-cos(x4+4)sin(x⁴+4)dx

Homework Equations

Let u = cos(x⁴+4)

The Attempt at a Solution

I know the answer does not have the sin in it and only the e remains, because when the integral is found e stays unchanged.
I could find somewhere online to calculate it, but I want to know how to do it, because questions like this will be on the final exam.

 

[Dick]

Homework Statement

∫8x³e^-cos(x4+4)sin(x⁴+4)dx

Homework Equations

Let u = cos(x⁴+4)

The Attempt at a Solution

I know the answer does not have the sin in it and only the e remains, because when the integral is found e stays unchanged.
I could find somewhere online to calculate it, but I want to know how to do it, because questions like this will be on the final exam.

You have the correct u substitution. Now use it. What's du?

 

[micromass]

So, what do you get after that u substitution??

 

[CaptainK]

So I've found du and put it into the form

∫udv = uv - ∫vdu

so for v I have 8x³

du = -sin(4x³) dx

dv = 24x² dx

Which gives me

∫cos(x⁴+4)24x²dx = cos(x⁴+4)8x³ - ∫8x³(-sin4x³) dx

But I feel like I'm going down the wrong path, especially since the e isn't present

 

[Dick]

You are going down the integration by parts path. It's not the right way to go. Why do you think du=(-sin(4x^3))dx?

 

[CaptainK]

So using substitution

du = sin(x⁴+4)

because cos converted to sin doesn't change its sign, I thought it did but its for converting from cos to sin.

t = u

∫f'(g(x))g'(x) dx = ∫f'(t) dt/dx = ∫ f'(t)dt = f(t) + C

= f(g(x)) + C when substituting back in

f(x) = original equation and
t = u = cos(x⁴+4)

 

[Dick]

So using substitution

du = sin(x⁴+4)

because cos converted to sin doesn't change its sign, I thought it did but its for converting from cos to sin.

t = u

∫f'(g(x))g'(x) dx = ∫f'(t) dt/dx = ∫ f'(t)dt = f(t) + C

= f(g(x)) + C when substituting back in

f(x) = original equation and
t = u = cos(x⁴+4)

df(x)=f'(x)*dx. f(x) here is cos(x^4+4). What is df(x)?

 

[CaptainK]

so f(x) = cos(x⁴+4)

then

f'(x) = sin(x⁴+4) + 4x³

which gives

∫cos(x⁴+4) +4x³ *sin(x⁴+4) + C

 

[Dick]

so f(x) = cos(x⁴+4)

then

f'(x) = sin(x⁴+4) + 4x³

which gives

∫cos(x⁴+4) +4x³ *sin(x⁴+4) + C

I'd suggest you review the chain rule. If f(x)=cos(x^4+4) then the derivative f'(x) is not what you wrote. Give it another try. Tell me what the chain rule says.

 

[CaptainK]

Chain Rule states

dy/dx = dy/du * dx/du

∫8x³e^-cos(x4+4)sin(x⁴+4)

8∫x³∫e^{-cos(x4+4)∫}sin(x⁴+4)

8∫x⁴/4∫e^-cos(x4+4)∫cos(x⁴+4)+4x³

 

[Dick]

Chain Rule states

dy/dx = dy/du * dx/du

∫8x³e^-cos(x4+4)sin(x⁴+4)

8∫x³∫e^{-cos(x4+4)∫}sin(x⁴+4)

8∫x⁴/4∫e^-cos(x4+4)∫cos(x⁴+4)+4x³

Stop worrying about the integral for a while. You are having trouble differentiating f(x)=cos(x^4+4). Try and get that right first.

 

[CaptainK]

Finding the derivative of cos(x⁴+4)

I got -4x³sin(x⁴+4)

Do I then follow the formula above with f(x) and t and then solve for the Integral.

 

[Dick]

Finding the derivative of cos(x⁴+4)

I got -4x³sin(x⁴+4)

Do I then follow the formula above with f(x) and t and then solve for the Integral.

Ok, you've got the derivative. If you take u=cos(x^4+4), that means du=(-4x^3)sin(x^4+4)dx. Now you should be able to turn the integral into
C \int e^{-u}du
where C is a numerical constant. This is u-substitution.