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Practice Problems for Module 4

Sections 7.1 and 7.2

Directions. The following are review problems. We recommend you work the problems yourself, and then click "Answer" to check your answer. If you do not understand a problem, you can click "Video" to learn how to solve it. You can also follow the link for the full video page to find related videos. 
  1. Evaluate the following integrals.
    1. \(\displaystyle \int \! x^2e^{2x} \, dx\)  ​

      \(\displaystyle \int \! x^2e^{2x} \, dx = \frac{1}{2}x^2e^{2x}-\frac{1}{2} xe^{2x} + \frac{1}{4}e^{2x}+C\)


      To see the full video page and find related videos, click the following link.
      WIR 20B M152 V22


    2. \(\displaystyle \int \! \left( \ln (x)\right)^2\, dx\)  ​

      \(\displaystyle \int \!\left( \ln (x)\right)^2\, dx = x\left( \ln (x)\right)^2-2x\ln(x)+2x+C\)


      To see the full video page and find related videos, click the following link.
      WIR 20B M152 V23


    3. \(\displaystyle \int \! x\tan^2(x)\, dx\)  ​

      \(\displaystyle \int \! x\tan^2(x)\, dx =x \tan(x)-\frac{1}{2} x^2+\ln|\cos(x)|+C\)


      To see the full video page and find related videos, click the following link.
      WIR 20B M152 V24


    4. \(\displaystyle \int \! 6x^5\cos(x^3) \, dx\)  ​

      \(\displaystyle \int \! 6x^5\cos(x^3) \, dx = 2x^3\sin(x^3)+2\cos(x^3)+C\)


      To see the full video page and find related videos, click the following link.
      WIR 20B M152 V25


    5. \(\displaystyle \int \! \sin(2x)e^{5x} \, dx\)  ​

      \(\displaystyle \int \! \sin(2x)e^{5x} \, dx = \frac{5}{29} \sin(2x)e^{5x}-\frac{2}{29} \cos(2x)e^{5x}+C\)


      To see the full video page and find related videos, click the following link.
      WIR 20B M152 V26


  2. Evaluate the following integrals.
    1. \(\displaystyle \int \! \tan^3(x) \, dx\)  ​

      \(\displaystyle \int \! \tan^3(x) \, dx =\frac{1}{2}\tan^2(x)+\ln|\cos(x)|+C\)


      To see the full video page and find related videos, click the following link.
      WIR 20B M152 V27


    2. \(\displaystyle \int \! \dfrac{\sin^3(x)}{\sec^2(x)} \, dx\)  ​

      \(\displaystyle \int \! \dfrac{\sin^3(x)}{\sec^2(x)} \, dx =\frac{1}{5}\cos^5(x)-\frac{1}{3}\cos^3(x)+C\)
      Video Errata: The rule of thumb mentioned in the video is not always true. Rather than worrying about the size of the powers when dealing with powers of trig functions time powers of trig functions, it’s better to consider which one can be reduced down in terms of the other after differentiating and then using trig identities. This often involves, but not always, the Pythagorean identities. You can then let \(u\) be the trig function you can write the other in terms of. For this problem, by letting \(u = \sin(x)\), you run into issues because you’ll have to deal with a single (not raised to a power) \(\cos(x)\) that’s left over when writing \(du\). You should try this yourself. However, sometimes either works just as well.



      To see the full video page and find related videos, click the following link.
      WIR 20B M152 V28


    3. \(\displaystyle \int \! \cos^2(x)\sin^2(x) \, dx\)  ​

      \(\displaystyle \int \! \cos^2(x)\sin^2(x) \, dx = \frac{x}{8} - \frac{1}{32}\sin(4x)+C\)


      To see the full video page and find related videos, click the following link.
      WIR 20B M152 V29


    4. \(\displaystyle \int \! \sec^4(x)\tan^3(x) \, dx\)  ​

      \(\displaystyle \int \! \sec^4(x)\tan^3(x) \, dx = \frac{1}{4}\tan^4(x)+\frac{1}{6}\tan^6(x)+C\)


      To see the full video page and find related videos, click the following link.
      WIR 20B M152 V30


    5. \(\displaystyle \int \! \sin(3x)\cos(x) \, dx\)  ​

      \(\displaystyle \int \! \sin(3x)\cos(x) \, dx = -\frac{1}{8}\cos(4x)-\frac{1}{4}\cos(2x)+C\)


      To see the full video page and find related videos, click the following link.
      WIR 20B M152 V31