$$ \because \frac d {dx}[g(f(x)]=g'(f(x))\cdot f'(x) \newline \therefore \int g'(f(x))f'(x)=g(f(x)) $$

Integration by Substitution

<aside> ⚙ Used for composite functions that is derivate with the chain rule….

</aside>

Looking at the integrands that consist of a product (sometimes quotient) of a function and its derivative [or something close to the derivative: wrong coefficient]

1. Determine which part of the integrand is function f(x). (set it to u)
2. Find dx (respect to x)
   1. Find the derivative of the function u.
   2. Use the derivative to isolate dx to one side.
      1. if $\frac {du} {dx} =Z$
      2. then $du=Z dx$, treat this as a fraction.
      3. at last $\frac {du} Z=dx$
3. Substitute dx back into the antiderivative.
   1. This can cancel derivative of the function u in the integrand
   2. if not, for a difference of coefficients
      1. multiple a constant to make the integrand to have one derivative of the one function they have.
      2. Also remember to counteract the changes (multiplication the constant that is mention in [i]) by the reciprocal.
4. treat the function as respect to u.
5. At last, set u back to the function.

RULE

Just remember if there is two functions that is seems to have been resulted after the chain rule, U substitution is always a good option.

Formula of Integration (reference)

<aside> ⚙ These are some of “rules” derivative from U substitution

</aside>

• $\int f'(x)(f(x))^n=\frac 1 {x+1} (f(x))^{n+1} +C, n\ne -1$

• $\int \frac {f'(x)} {f(x)} =\ln |f(x)|+C$

• $\int f'(x) e^{f(x)} =e^{f(x)}+C$

• $\int f'(x)a^{f(x)}=\frac 1 {\ln a} a^{f(x)}+C$

  ---

• $\int [f'(x)]\cos (f(x)) dx=\sin f(x)+C$

• $\int [f'(x)]\sec ^2 (f(x)) dx=\tan f(x)+ C$

• $\int [f'(x)]\sec (f(x)) \tan (f(x)) dx=\sec (f(x))+C$

• $\int [f'(x)] \sin (f(x))dx=-\cos (f(x))+C$

• $\int [f'(x) ]csc^2(f(x)) dx=-\cot (f(x))+C$

• $\int [f'x]\csc (f(x) )\cot (f(x)) dx= -\csc (f(x)) +C$