7.4: Homework- u-substitution

1. As a review of the chain rule from derivatives, find $\frac{d}{dx} e^{-x^2 + 1}$.
   $2x e^{x^2 + 1}$

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2. Read through section 6A again and then read through section 6B.
3. Watch the Khan Academy video u-substitution.
4. Compute the following indefinite integral using the method of $u$-substitution.

   $\int (5x^4 + 2x) e^{x^5 + x^2}dx$

   $e^{x^5 + x^2} + C$

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5. Watch another example of $u$-substitution: u-substitution 2.
6. Compute the following indefinite integral using the method of $u$-substitution.

   $\int \frac{2x}{x^2 - 5}dx$

   $\ln(x^2 - 5) + C$

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7. Reread the part about the chain rule shortcut for $u$-substitution in chapter 6 of the online notes, and reread Example 6B.2. Then try the following problems.
   1. $\int e^{-3x}dx$.
      $\frac{1}{-3} e^{-3x} + C$

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   2. $\int \left( \frac{1}{2}x - 1 \right)^4dx$
      $.4 \left( \frac{1}{2} x - 1 \right)^5 + C$

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      .

   3. $\int_0^2 (5x - 3)^3dx$.
      $116$

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   4. $\int \frac{1}{7x - 2}dx$
      $\frac{1}{7} \ln(7x - 2) + C$

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   5. $\int \frac{1}{\sqrt{0.5x - 1}}dx$
      $4 \sqrt{0.5x - 1} + C$

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8. Try some more $u$-substitution integrals.
   1. $\int (8x^3) (x^4 + 1)^2dx$
      1. We see that $u = x^4 + 1$ in this case.
      2. We see $\frac{du}{dx} = 4x^3$, so $dx = \frac{1}{4x^3} du$.
      3. We have

         $$\begin{align*} & = \int (8x^3) u^2 \frac{1}{4x^3} du \\ & = \int 8 u^2 \frac{1}{4}du \\ & = 2 \int u^2du \end{align*}$$

      4. Integrating we have

         $2 \int u^2du = 2 \left( \frac{1}{3} u^3 \right ) + C = \frac{2}{3} u^3 + C$

      5. Substitution of $u = x^4 + 1$ yields

         $\frac{2}{3} (x^4 + 1)^3 + C$

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   2. $\int (x^4 + 1)^2 (8x^3)dx$
      This is exactly the same as the previous problem, just written a different way. No need to redo work.

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   3. $\int (3x^2 - 1) e^{x^3 - x}dx$
      1. We see that $u = x^3 - x$ in this case.
      2. We see $\frac{du}{dx} = 3x^2 - 1$, so $dx = \frac{1}{3x^2 - 1} du$.
      3. We have

         $$\begin{align*} & = \int (3x^2 - 1) e^u \frac{1}{3x^2 - 1} du \\ & = \int e^udu \end{align*}$$

      4. Integrating we have

         $\int e^udu = e^u + C$

      5. Substitution of $u = x^3 - x$ yields

         $e^{x^3 - x} + C$

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   4. $\int (x^2 - \frac{1}{3}) e^{x^3 - x}dx$
      $\frac{1}{3} e^{x^3 - x} + C$

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   5. $\int (e^x+ x)^5 (e^x+1)dx$
      $\frac{1}{6} (e^x + x)^6 + C$

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   6. $\int \frac{2x}{x^2 - 1}dx$
      $\ln(x^2 -1) + C$

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   7. $\int e^x \sqrt{e^x + 1}dx$
      

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   8. $\int \frac{\sin(x)}{\cos(x)}dx$
      $-\ln(\cos(x)) + C$

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   9. $\int \frac{\sqrt{ \ln(x) }}{x}dx$
      

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   10. $\int \frac{x}{(x^2 - 5)^3}dx$
       1. We see that $u = x^2 - 5$ in this case.
       2. We see $\frac{du}{dx} = 2x$, so $dx = \frac{1}{2x} du$.
       3. We have

          $$\begin{align*} & = \int \frac{x}{u^3}\frac{1}{2x}du \\ & = \int \frac{1}{2u^3}du \\ & = \int \frac{1}{2}u^{-3}du \end{align*}$$

       4. Integrating we have

          $\int \frac{1}{2}u^{-3}du = - \frac{1}{4} u^{-2} + C$

       5. Substitution of $u = x^2 - 5$ yields

          $- \frac{1}{4} (x^2 - 5)^{-2} + C$

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   11. $\int \frac{\cos(x)}{\sqrt{\sin(x) + 1}}dx$
       $2\sqrt{\sin(x) + 1} + C$

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   12. $\int \frac{1}{x (2 \ln(x) + 1)^4}dx$.
       $-\frac{1}{6(2 \ln(x) + 1)^3} + C$

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   13. $\int (e^{5x} + 1)^9 (e^{5x} )dx$.
       $\frac{1}{2} (e^{5x} + 1)^{10} + C$

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