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Practice Problems for 5/1

  1. Evaluate the following indefinite integrals.
    1. \(\int \frac{\sec^2(x)}{(5-\tan x)^7}\, dx \)
      2. Let \(u=5-\tan x\). Then \(\frac{du}{dx}=-\sec^2x\) and making a substitution of variables gives \[ \int \frac{\sec^2(x)}{(5-\tan x)^7}\, dx=\int -\frac{1}{u^7}\,du=\frac{1}{6u^6}+C=\frac{1}{6(5-\tan x)^6}+C \]
    2. \(\int 9xe^{5-3x^2}\, dx\)
      3. We see a composition where the inside function is \(5-3x^2\). Let this define a new variable \(u\). Then \(\frac{du}{dx}=-6x=\frac{-2}{3}9x\), and \[ \int 9xe^{5-3x^2}\, dx=\int \frac{-3}{2}e^u\,du=\frac{-3}{2}e^u+C=\frac{-3}{2}e^{5-3x^2}+C. \] To check that we multiplied by the correct constant, take the derivative of the answer to check: \[ \frac{d}{dx}\frac{-3}{2}e^{5-3x^2}+C=\frac{-3}{2}(-6x)e^{5-3x^2}=9xe^{5-3x^2}, \] which is the original integrand, so this is correct.

  2. Evaluate the following definite integrals.
    1. \(\int_{-1}^3 (2+6x^2)(x+x^3)^{3/4} \, dx \)
      2. Let \(u=x+x^3\), so \(\frac{du}{dx}=1+3x^2\). Also, for this substitution of variables, \(u(-1)=-2\) and \(u(3)=30\), so the integral is \[ \int_{-1}^3 (2+6x^2)(x+x^3)^{3/4} \, dx=\int_{-2}^{30}2u^{3/4}\,du=\frac{8}{7}u^{7/4}|_{-2}^{30}=\frac{8}{7}[30^{7/4}-(-2)^{7/4}]. \]
    2. \(\int_0^{\ln(\pi-2)} e^x\cos(2+e^x)\, dx\)
      3. Let \(u=2+e^x\). Then \(\frac{du}{dx}=e^x\), \(u(0)=2+1=3\), and \(u(\ln(\pi-2))=2+(\pi-2)=\pi\). Making the substitution gives \[ \int_0^{\ln(\pi-2)} e^x\cos(2+e^x)\, dx=\int_3^{\pi} \cos u\, du=\sin(\pi)-\sin(3)=-\sin(3). \]

  3. Find the average value of \(f(x)=x(5+x^2)^2\) on the interval \([0,5]\).
    4. \[ f_{\textrm{avg}}=\frac{1}{5}\int_0^5 x(5+x^2)^2\, dx \] To find the value of the integral, note one of the terms of the integrand is a composition. We therefore make the substitution \(u=5+x^2\), so that \[ f_{\textrm{avg}}=\frac{1}{5}\int_0^5 x(5+x^2)^2\, dx=\frac{1}{5}\int_5^{30}\frac{1}{2}u^2\,du=\frac{1}{30}(30^3-5^3). \]