Week 14 - Substitution Rule #

Theorem #

• If $u = g(x)$ is a differentiable function whose range is an interval $I$ and $f$ is continuous on $I$, then $$\int f(g(x))g’(x)dx = \int f(u) du$$
  • Like reverse the chain rule.
  • It is permissible to operate with dx and du after integral signs as if they were differentials.

Examples #

• $\int\frac{ x }{ \sqrt{ 4-9x^2 } }dx$
  • $$\begin{aligned} u &= 4-9x^2, du = -18x \\ \int\frac{ x }{ \sqrt{ 4-9x^2 } } dx &= -\frac{1}{18} \int \frac{ -18x }{ \sqrt{4-9x^2 } } \\ &= -\frac{1}{18} \int \frac{ du }{ \sqrt{ u } } \\ &= -\frac{1}{18} \frac{ u^{ 1/2 } }{ 1/2 } + C \\ &= -\frac{1}{9} \sqrt{4-9x^2} + C \end{aligned}$$
• $\int\frac{ 1 }{ \sqrt{ 4-9x^2 } }dx$
  • $$\begin{aligned} \int\frac{1}{ \sqrt{ 4-9x^2 } }dx &= \int\frac{1}{2\sqrt{ 1-(\frac{ 3 }{ 2 } x)^2 } }dx \\ u &= \frac{3}{2}x, du = \frac{3}{2} \\ \int\frac{ 1 }{ \sqrt{ 4-9x^2 } } dx &= \frac{1}{3}\int \frac{1}{ \sqrt{ 1-u^2 } }du \\ &= \frac{1}{3} \arcsin{u} + C \\ &= \frac{1}{3} \arcsin{\frac{ 3 }{ 2 } x } + C \end{aligned}$$

Handle the Endpoints #

• Normally, we’ll do $$\int_a^bf’(g(x))g’(x)dx = f(g(x))\big]_a^b$$
• Change the endpoints with u $$\int_a^bf’(g(x))g’(x)dx = f(u)\big]_{g(a)}^{g(b)}$$

Example #

• $\int_{x=0}^2 2x(x^2+1)^3 dx$
• set $u = x^2+1$, we get $\int_{x=0}^2 u^3 du$
• In the normal way, $\int_{x=0}^2 u^3 du = \frac{u^4}{4}\big]{x=0}^2 = \frac{(x^2+1)^4}{4}\big]{x=0}^2 = \frac{(2^2+1)^4}{4} - \frac{(0^2+1)^4}{4} = 156$.
• Another way, change the endpoints. $\int_{x=0}^2 u^3 du = \int_{u=1}^5 u^3 du = \frac{u^4}{4}\big]_{u=1}^5 = \frac{5^4}{4} - \frac{1^4}{4} = 156$.

Do U-substitution More Than Once #

• $\int -2 \cos{x} \sin{x} \cos{(\cos^2{x}+1)} dx$
  • set $u = \cos{x}, du = -\sin{x} dx$
  • = $\int 2u \cos{(u^2+1)} du$
  • set $v = u^2+1, dv = 2u$
  • = $\int \cos{v} dv$
  • = $\sin{v} + C$
  • = $\sin{u^2+1} + C$
  • = $\sin{(\cos^2{x}+1)} + C$

Some Tricks #

1. use $\arctan{x} = \frac{1}{x^2+1}$
   • $\int \frac{1}{x^2+4x+7} dx = \int \frac{1}{(x+2)^2+3} dx = \int \frac{1}{(x+2)^2+3} dx = \frac{1}{3} \int \frac{1}{(\frac{1}{ \sqrt{ 3 } }(x+2))^2+1} dx$
     • set $u = \frac{1}{ \sqrt{ 3 } }(x+2), u’ = \frac{1}{\sqrt{3} }$
     • $= \frac{1}{\sqrt{3} } \int \frac{1}{u^2+1} du = \frac{1}{\sqrt{3} }\arctan(u) + C = \frac{1}{\sqrt{3} }\arctan(\frac{1}{ \sqrt{ 3 } }(x+2)) + C$
2. rationalizing substitution
   • $\int \frac{x}{\sqrt[3]{x+1} }$
     • Take the whole denominator as $u$: $u = \sqrt[3]{x+1}$, then $x = u^3 - 1$
     • $= \int \frac{u^3-1}{u}du$
3. multiple a reciprocal to make the substitution visible.
   • $\int \frac{1}{1+\cos{x} } dx$
     • $= \int \frac{1}{1+\cos{x} } \cdot \frac{1-\cos{x} }{1+\cos{x} } dx$
     • $= \int \frac{1-\cos{x} }{\sin^2{x} } dx = \int \frac{1}{\sin^2{x} } dx - \int \frac{\cos{x} }{\sin^2{x} } dx = - \cot{x} - \int \frac{\cos{x} }{\sin^2{x} } dx$
     • set $u = \sin{x}, u’ = \cos{x}$
     • $= - \cot{x} - \int \frac{1}{u^2} du = - \cot{x} + \frac{1}{u} + C = - \cot{x} + \csc{x} + C$

Differentiate Integral Functions #

• Base on the limit theorem, we know that at point x, the change of the integral function should be f(x). But what if the upper endpoint is a function?
• Like $\frac{d}{dx}\int_0^{g(x)} f(x) dx$.
• Use the chain rule: $\frac{d}{dx}f(g(x)) = f’(g(x)) \cdot g’(x)$ and the Fundamental Theorem: $F’(x) = f(x)$, we get:
  • $\frac{d}{dx}\int_0^{g(x)} f(x) dx = F’(g(x)) \cdot g’(x) = f(g(x)) \cdot g’(x)$
• For example: $\frac{d}{dx}\int_0^{x^2} \sin{t} dt$, $g(x) = x^2$
  • $\frac{d}{dx} \int_0^{x^2} \sin{x} dt = \sin{x^2} \cdot 2x$

Use The Extreme Value Theorem to Prove Fundamental Theorem #

• We want to do two things. First to prove the limit exits, and second, find the value of that derivative: $$F’(x) = \lim_{h \to 0^{+} } \frac{F(x+h) - F(x)}{h} = f(x)$$
  • to make it a little easier, we only consider h is positive.
  • $F’(x) = \lim_{h \to 0^{+} } \frac{\int_a^{x+h} f(t) dt - \int_a^{x} f(t) dt}{h}$
  • = $\lim_{h \to 0^{+} } \frac{\int_x^{x+h} f(t) dt}{h}$
  • Base on the The Extreme Value Theorem, we know that there must be a:
    • $m(h)$ = min value of f on interval $[x, x+h]$
    • $M(h)$ = max value of f on interval $[x, x+h]$
    • $m(h) \cdot h \le \int_x^{x+h} f(t) dt \le M(h) \cdot h$
    • $m(h) \le \frac{\int_x^{x+h} f(t) dt}{h} \le M(h)$
    • Use the Squeeze Theorem, we know that:
      • $\lim_{h \to 0^+}m(h) = f(x) = lim_{h \to 0+}M(h)$
      • Then $\lim_{h \to 0^+}\frac{\int_x^{x+h} f(t) dt}{h} = f(x)$
    • So $F’(x) = f(x)$