I need Step-By-Step Help

orcsmoocher

Okay, so I've sat through the lecture, took notes, and followed along with some integration-by-parts and substitution videos and I still don't get it. It just makes 0 sense. Usually I also try and follow the steps of a few of those step-by-step solvers, but Symbolab paywalls the solution and Wolfram Alpha (which I paid for on my phone) basically skips ALL of the steps. I'm frustrated. Please show me how I'd do this, step-by-step and I'll be in your debt forever. Integration is the worst.

topsquark

Forum Staff
I'm going to assume that z is a real variable and not a complex number.

You can take the derivative of just about anything but integration tends to be an art form.

One of the first things to do when you see a trigonometric integration is to see of you can either set u = sin(x) or u = cos(x). I'd suggest u = sin(x). What does the integral look like when you do this?

-Dan

Plato

MHF Helper
The simple primitive is $$\displaystyle \int {\frac{{\cos (x)}}{{{{\sin }^4}(x)}}dx = \frac{{ - 1}}{3}{{\csc }^3}(x)} + C$$
By differentiating that result and reverse engineering you should see the steps.

topsquark

orcsmoocher

I'm going to assume that z is a real variable and not a complex number.

You can take the derivative of just about anything but integration tends to be an art form.

One of the first things to do when you see a trigonometric integration is to see of you can either set u = sin(x) or u = cos(x). I'd suggest u = sin(x). What does the integral look like when you do this?

-Dan
(yes, z is just the variable used instead of x)

Would you set u = sin(x) or sin^4(x)? What would v be?

orcsmoocher

The simple primitive is $$\displaystyle \int {\frac{{\cos (x)}}{{{{\sin }^4}(x)}}dx = \frac{{ - 1}}{3}{{\csc }^3}(x)} + C$$
By differentiating that result and reverse engineering you should see the steps.
I'm sorry, I don't understand...

Debsta

MHF Helper
Let u = sin(z). There is no v involved. This is integration by substitution NOT integration by parts.
I'll walk you through it.
What is du/dz?

(Note: i have limited time today, so please respond asap)

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topsquark

Plato

MHF Helper
I'm sorry, I don't understand...
You are to find the derivative of $$\displaystyle {\frac{{ - 1}}{3}{{\csc }^3}(x)}$$ from which you show the derivative is equivalent to $$\displaystyle {\frac{{\cos (x)}}{{{{\sin }^4}(x)}}}$$

topsquark

topsquark

Forum Staff
(yes, z is just the variable used instead of x)

Would you set u = sin(x) or sin^4(x)? What would v be?
Try u = sin(z). And this is simple substitution... no v required. (Integration by parts would not be a good idea for this one.)

-Dan

orcsmoocher

Let u = sin(z). There is no v involved. This is integration by substitution NOT integration by parts.
I'll walk you through it.
What is du/dz?

(Note: i have limited time today, so please respond asap)
Okay, if u = sin(z) then du = -cos(z), right?

orcsmoocher

Try u = sin(z). And this is simple substitution... no v required. (Integration by parts would not be a good idea for this one.)

-Dan
How do I tell when to use substitution or integration by parts?