INTEGRATION BY U-SUBSTITUTION
Lets go back to CHAIN RULE
if we have an equation such as
y = f(g(x))
then we know that to differentiate this equation we should apply
chain rule
then
d
f(g(x)) = f'(g(x))g'(x)
dx
or we can let u = g(x)
d d
then f(g(x)) = f'(u) u
dx du
thus = f'(g(x))g'(x)
Then the integral of f'(g(x))g'(x) can be found using u - substitution
integral means going back to the original equation or function
if f'(x) then the ∫f'(x)dx = f(x) + c
{where c represents any constant number}
d
we know that f(g(x)) = f'(g(x))g'(x) then lets prove that the
dx
∫f'(g(x))g'(x)dx = f(g(x))
we going to use u - substitution
therefore
∫f'(g(x))g'(x)dx
first we let u = g(x)
then derive
du = g'(x)dx
Lets go back to CHAIN RULE
if we have an equation such as
y = f(g(x))
then we know that to differentiate this equation we should apply
chain rule
then
d
f(g(x)) = f'(g(x))g'(x)
dx
or we can let u = g(x)
d d
then f(g(x)) = f'(u) u
dx du
thus = f'(g(x))g'(x)
Then the integral of f'(g(x))g'(x) can be found using u - substitution
integral means going back to the original equation or function
if f'(x) then the ∫f'(x)dx = f(x) + c
{where c represents any constant number}
d
we know that f(g(x)) = f'(g(x))g'(x) then lets prove that the
dx
∫f'(g(x))g'(x)dx = f(g(x))
we going to use u - substitution
therefore
∫f'(g(x))g'(x)dx
first we let u = g(x)
then derive
du = g'(x)dx
