Integration by Substitution

Theorem. If $x=\varphi(t)$ is a differentiable function of $t$, then $\int f(x)\,dx=\int f[\varphi(t)]\,\varphi'(t)\,dt$.

Reasoning: since $x=\varphi(t)$, $\frac{dx}{dt}=\varphi'(t)$. If $\int f(x)\,dx=g(x)$, then by the chain rule $\frac{d}{dt}[g(x)]=\frac{d}{dx}[g(x)]\cdot\frac{dx}{dt}=f(x)\cdot\varphi'(t)=f[\varphi(t)]\varphi'(t)$. So $g(x)$ is also a primitive, with respect to $t$, of $f[\varphi(t)]\varphi'(t)$; that is exactly the claimed identity, read backwards.

Illustration: $\int 3x^2\sin(x^3)\,dx$. Put $x^3=t$, so $3x^2\,dx=dt$; the integral becomes $\int\sin t\,dt=-\cos t+c=-\cos(x^3)+c$.

Four corollaries of this theorem cover the recurring substitution patterns.

Corollary I. If $\int f(x)\,dx=g(x)+c$ then $\int f(ax+b)\,dx=g(ax+b)\cdot\frac1a+c$. Proof: put $ax+b=t$, so $dx=\frac1a\,dt$; then $\int f(ax+b)\,dx=\frac1a\int f(t)\,dt=\frac1a g(t)+c=\frac1a g(ax+b)+c$. Example: $\int\sec^2(5x-4)\,dx=\frac15\tan(5x-4)+c$.

Corollary II. $\int[f(x)]^n f'(x)\,dx=\dfrac{[f(x)]^{n+1}}{n+1}+c,\ n\neq-1$. Proof: put $f(x)=t$, so $f'(x)\,dx=dt$; then $\int t^n\,dt=\frac{t^{n+1}}{n+1}+c$. Example: $\int\dfrac{(\sin^{-1}x)^3}{\sqrt{1-x^2}}\,dx=\int(\sin^{-1}x)^3\cdot\frac{1}{\sqrt{1-x^2}}\,dx=\dfrac{(\sin^{-1}x)^4}{4}+c$, using $f(x)=\sin^{-1}x$.

Corollary III. $\int\dfrac{f'(x)}{f(x)}\,dx=\log|f(x)|+c$. Proof: put $f(x)=t$, so $f'(x)\,dx=dt$; then $\int\frac1t\,dt=\log|t|+c$. Example: $\int\cot x\,dx=\int\dfrac{\cos x}{\sin x}\,dx=\log|\sin x|+c$, since $\frac{d}{dx}\sin x=\cos x$. …