$$ ∫udv=(u)(v)-∫vdu $$

• Where u = your choice of u sub and dv= the rest of the function
  • u should be set to something when you take it’s derivative it shows up in the problem or makes it easier to evaluate as a new integral
• v will then be the antiderivative of dv and du will be the derivative of u

General Notes:

• You can use integration by parts as many times as needed, and often you will have to do it more than once
• Always remember to write the original ∫udv = whatever you found in doing by parts, because sometimes your integral expression on the right can be added/subtracted to the original problem to get the final answer
• Usually, we can to make v something easier to integrate as it will become the integrand after using LIATE (Log Inverse Algebraic Trig Exponential)

Example:

$$ ∫sin(x)cos(x)dx $$

1. let u=sin(x) then dv=cos(x)dx
2. therefore we get:

$$ (sin(x))(-sin(x))-∫-sin(x)cos(x) $$

1. Taking out the negative gets you:

$$ (sin(x))(-sin(x))+∫sin(x)cos(x) $$

1. We can now add this new integral to our original, as they are the same:

$$ 2∫sin(x)cos(x)=(sin(x))(-sin(x))=sin^2(x)/2 $$

I just added the right same integral to our original to get 2 of them, then solving you divide by 2 and multiply the sin together which gets you $sin^2(x)$ all over 2