Week 14 - Substitution Rule

1. Theorem

  • If u=g(x)u = g(x) is a differentiable function whose range is an interval II and ff is continuous on II, then f(g(x))g(x)dx=f(u)du\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.

1.1. Examples

  • x49x2dx\int\frac{ x }{ \sqrt{ 4-9x^2 } }dx
    • u=49x2,du=18xx49x2dx=11818x49x2=118duu=118u1/21/2+C=1949x2+C\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}
  • 149x2dx\int\frac{ 1 }{ \sqrt{ 4-9x^2 } }dx
    • 149x2dx=121(32x)2dxu=32x,du=32149x2dx=1311u2du=13arcsinu+C=13arcsin32x+C\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}

2. Handle the Endpoints

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

2.1. Example

  • x=022x(x2+1)3dx\int_{x=0}^2 2x(x^2+1)^3 dx
  • set u=x2+1u = x^2+1, we get x=02u3du\int_{x=0}^2 u^3 du
  • In the normal way, x=02u3du=u44]x=02=(x2+1)44]x=02=(22+1)44(02+1)44=156\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. x=02u3du=u=17u3du=u44]u=17=744144=156\int_{x=0}^2 u^3 du = \int_{u=1}^7 u^3 du = \frac{u^4}{4}\big]_{u=1}^7 = \frac{7^4}{4} - \frac{1^4}{4} = 156.

3. Do U-substitution More Than Once

  • 2cosxsinxcos(cos2x+1)dx\int -2 \cos{x} \sin{x} \cos{(\cos^2{x}+1)} dx
    • set u=cosx,du=sinxdxu = \cos{x}, du = -\sin{x} dx
    • = 2ucos(u2+1)du\int 2u \cos{(u^2+1)} du
    • set v=u2+1,dv=2uv = u^2+1, dv = 2u
    • = cosvdv\int \cos{v} dv
    • = sinv+C\sin{v} + C
    • = sinu2+1+C\sin{u^2+1} + C
    • = sin(cos2x+1)+C\sin{(\cos^2{x}+1)} + C

4. Some Tricks

  1. use arctanx=1x2+1\arctan{x} = \frac{1}{x^2+1}
    • 1x2+4x+7dx=1(x+2)2+3dx=1(x+2)2+3dx=131(13(x+2))2+1dx\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=13(x+2),u=13u = \frac{1}{ \sqrt{ 3 } }(x+2), u' = \frac{1}{\sqrt{3} }
      • =131u2+1du=13arctan(u)+C=13arctan(13(x+2))+C= \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
    • xx+13\int \frac{x}{\sqrt[3]{x+1} }
      • Take the whole denominator as uu: u=x+13u = \sqrt[3]{x+1}, then x=u31x = u^3 - 1
      • =u31udu = \int \frac{u^3-1}{u}du
  3. multiple a reciprocal to make the substitution visible.
    • 11+cosxdx\int \frac{1}{1+\cos{x} } dx
      • =11+cosx1cosx1+cosxdx= \int \frac{1}{1+\cos{x} } \cdot \frac{1-\cos{x} }{1+\cos{x} } dx
      • =1cosxsin2xdx=1sin2xdxcosxsin2xdx=cotxcosxsin2xdx= \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=sinx,u=cosxu = \sin{x}, u' = \cos{x}
      • =cotx1u2du=cotx+1u+C=cotx+cscx+C= - \cot{x} - \int \frac{1}{u^2} du = - \cot{x} + \frac{1}{u} + C = - \cot{x} + \csc{x} + C

5. 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 ddx0g(x)f(x)dx\frac{d}{dx}\int_0^{g(x)} f(x) dx.
  • Use the chain rule: ddxf(g(x))=f(g(x))g(x)\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x) and the Fundamental Theorem: F(x)=f(x)F'(x) = f(x), we get:
    • ddx0g(x)f(x)dx=F(g(x))g(x)=f(g(x))g(x)\frac{d}{dx}\int_0^{g(x)} f(x) dx = F'(g(x)) \cdot g'(x) = f(g(x)) \cdot g'(x)
  • For example: ddx0x2sintdt\frac{d}{dx}\int_0^{x^2} \sin{t} dt, g(x)=x2g(x) = x^2
    • ddx0x2sinxdt=sinx22x\frac{d}{dx} \int_0^{x^2} \sin{x} dt = \sin{x^2} \cdot 2x

6. 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)=limh0+F(x+h)F(x)h=f(x)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)=limh0+ax+hf(t)dtaxf(t)dthF'(x) = \lim_{h \to 0^{+} } \frac{\int_a^{x+h} f(t) dt - \int_a^{x} f(t) dt}{h}
    • = limh0+xx+hf(t)dth\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)m(h) = min value of f on interval [x,x+h][x, x+h]
      • M(h)M(h) = max value of f on interval [x,x+h][x, x+h]
      • m(h)hxx+hf(t)dtM(h)hm(h) \cdot h \le \int_x^{x+h} f(t) dt \le M(h) \cdot h
      • m(h)xx+hf(t)dthM(h)m(h) \le \frac{\int_x^{x+h} f(t) dt}{h} \le M(h)
      • Use the Squeeze Theorem, we know that:
        • limh0+m(h)=f(x)=limh0+M(h)\lim_{h \to 0^+}m(h) = f(x) = lim_{h \to 0+}M(h)
        • Then limh0+xx+hf(t)dth=f(x)\lim_{h \to 0^+}\frac{\int_x^{x+h} f(t) dt}{h} = f(x)
      • So F(x)=f(x)F'(x) = f(x)

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