Monday, August 23, 2010 - 14:00 - Tuesday, August 24, 2010 - 15:00
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Math 2300 Section 005 – Calculus II – Fall 2010

Substitution Examples – August 23–24, 2010

Guess and Check & Mechanical Substitution Examples

  1. \displaystyle\int x\cos(x^{2})\, dx

    Solution: If you guess that \sin(x^{2}) is an appropriate antiderivative, then you differentiate to 2x\cos(x^{2}) via the chain rule. But, we're now off by a scalar factor, which is an easy fix. Just multiply your antiderivative by half. We end up with

    \int x\cos(x^{2})\, dx=\frac{1}{2}\sin(x^{2})+C

    and we know this is the answer because we can differentiate to x\cos(x^{2}).

  2. \displaystyle\int t^{2}e^{{5t^{3}}}\, dt

    Solution: Let w=5t^{3} so that dw=15t^{2}\, dt. Then

    \int t^{2}e^{{5t^{3}}}\, dt\ =\ \frac{1}{15}\int e^{w}\, dw\ =\ \frac{1}{15}e^{w}+C\ =\ \frac{1}{15}e^{{5t^{3}}}+C.
  3. \displaystyle\int\cos x\sqrt{\sin x+1}\, dx

    Solution: Letting w=\sin x+1 we obtain dw=\cos x\, dx and so

    \displaystyle\displaystyle\int\cos x\sqrt{\sin x+1}\, dx \displaystyle=\int\sqrt{w}\, dw
    \displaystyle=\int w^{{1/2}}\, dw
    \displaystyle=\frac{2}{3}(\sin x+1)^{{3/2}}+C.
  4. \displaystyle\int x^{2}e^{{x^{3}+1}}\, dx

    Solution: Let w=x^{3}+1, so that dw=3x^{2}\, dx. Then we have \displaystyle\frac{dw}{3}=x^{2}\, dx and

    \displaystyle\int x^{2}e^{{x^{3}+1}}\, dx \displaystyle=\int e^{w}\frac{dw}{3}
    \displaystyle=\frac{1}{3}\int e^{w}\, dw
  5. \displaystyle\int\sqrt{\cos 3t}\,\sin 3t\, dt

    Solution: Let w=\cos 3t, so that dw=-3\sin(3t)\, dt (by the chain rule) and \displaystyle-\frac{dw}{3}=\sin(3t)\, dt. Then

    \displaystyle\int\sqrt{\cos 3t}\,\sin 3t\, dt \displaystyle=\int-\sqrt{w}\frac{dw}{3}
    \displaystyle=-\frac{1}{3}\int\sqrt{w}\, dw
    \displaystyle=-\frac{1}{3}\int w^{{1/2}}\, dw
    \displaystyle=-\frac{2}{9}(\cos 3t)^{{3/2}}+c
  6. \displaystyle\int\tan\theta\, d\theta

    Solution: Since \tan\theta=\sin\theta/\cos\theta, we set w=\cos\theta and we have dw=-\sin\theta\, d\theta. We then have

    \displaystyle\int\tan\theta\, d\theta \displaystyle=\int\frac{\sin\theta}{\cos\theta}\, d\theta
    \displaystyle=-\int\frac{1}{w}\, dw

Challenging Substitution Examples

  1. \displaystyle\int x\sqrt{1+x}\, dx

    Solutions: If we make the substitution u=1+x then we have du=dx. This doesn't seem to help get rid of the remaining x term in the integral, but notice that since u=1+x we can solve for x and write u-1=x. Thus, we find

    \displaystyle\int x\sqrt{1+x}\, dx \displaystyle=\int(u-1)\sqrt{u}\, du
    \displaystyle=\int u\sqrt{u}-\sqrt{u}\, du
    \displaystyle=\int u^{{3/2}}-u^{{1/2}}\, du
  2. \displaystyle\int\frac{x}{1+x^{4}}\, dx

    Solution: This example isn't immediately obvious at all. (Congratulations to those of you who spotted the path to the solution in class!) If we rewrite the integral and use w=x^{2} (and so dw=2x\, dx) then we have

    \displaystyle\int\frac{x}{1+x^{4}}\, dx \displaystyle=\int\frac{x}{1+(x^{2})^{2}}\, dx
    \displaystyle=\frac{1}{2}\int\frac{1}{1+w^{2}}\, dw

Definite Substitution Examples

  1. \displaystyle\int _{0}^{{1/2}}\cos(\pi x)\, dx

    Solution: Making the substitution w=\pi x and dw=\pi dx we find (in this case we do not change the limits of integration – they stay in terms of x)

    \displaystyle\int _{0}^{{1/2}}\cos(\pi x)\, dx \displaystyle=\int _{{x=0}}^{{x=1/2}}\cos w\frac{dw}{\pi}
    \displaystyle=\frac{1}{\pi}\int _{{x=0}}^{{x=1/2}}\cos w\, dw
    \displaystyle=\frac{1}{\pi}\sin w\Big|_{{x=0}}^{{x=1/2}}
    \displaystyle=\frac{1}{\pi}\sin\pi x\Big|_{{x=0}}^{{x=1/2}}
    \displaystyle=\frac{1}{\pi}\left(\sin(\pi/2)-\sin 0\right)
  2. \displaystyle\int _{0}^{2}\frac{x}{(1+x^{2})^{2}}\, dx

    Solution: Using w=1+x^{2}, and dw=2x\, dx, we have (here we change the limits of integration to be in terms of w)

    \displaystyle\int _{0}^{2}\frac{x}{(1+x^{2})^{2}}\, dx \displaystyle=\int _{{w=1}}^{{w=5}}\frac{x}{w^{2}}\, dx
    \displaystyle=\frac{1}{2}\int _{{w=1}}^{{w=5}}w^{{-2}}\, dw

© 2011 Jason B. Hill. All Rights Reserved.