Question

how to use u-substitution for integration?

Answer 1

Basically you find something that looks like it has its derivative in the integral in an attempt to reverse the chain rule easier.

Answer 2

similar to using systems in algebra to solve for an equation.
The rule is:
\[\int\limits f(g(x))g'(x)dx = \int\limits f(u)du\]
where u = g(x)

Answer 3

For example:
\[\int\limits \left( 1-\frac{ 1 }{ w }\right)\cos(w-\ln(w))dw\]

Answer 4

btw the derivative of ln(w) is 1/w

Answer 5

You should know how to integrate just a cosine so let’s make the substitution the stuff that is inside the cosine to make it simpler
\[u = w-\ln(w)\] \[du = (1-\frac{ 1 }{ w })dw\]
in front of the cosine appears exactly in the differential. The integral is then:
\[\int\limits (1-\frac{ 1 }{ w })\cos(w-\ln(w))dw = \int\limits \cos(u)dt\]
\[= \sin(u) + C\]
PLUG IT BACK IN
\[\sin(w-\ln(w))+C\]

Answer 6

Or a little simpler:
\[\int\limits \frac{\ln(x) dx}{x}; u=\ln(x) \implies \frac{du}{dx}=\frac{1}{x} \implies du=\frac{dx}{x}; \implies \]\[\int\limits u du=\frac{u^2}{2}+C=\frac{\ln^2(x)}{x}+C\]