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Mathematics LibreTexts

5.5E & 5.6E U-Substitution Exercises

  • Page ID
    14694
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    5.5: Substitution

    In the following exercises, find the antiderivative.

    261) \(\displaystyle∫(x+1)^4\,dx\)

    Answer:
    \(\displaystyle∫(x+1)^4\,dx \quad\) \(=\quad \displaystyle\frac{1}{5}(x+1)^5+C\)

    262) \(\displaystyle∫(x−1)^5\,dx\)

    263) \(\displaystyle∫(2x−3)^{−7}\,dx\)

    Answer:
    \(\displaystyle∫(2x−3)^{−7}\,dx\quad\) \(=\quad\displaystyle−\frac{1}{12(3−2x)^6}+C\)

    264) \(\displaystyle∫(3x−2)^{−11}\,dx\)

    265) \(\displaystyle∫\frac{x}{\sqrt{x^2+1}}\,dx\)

    Answer:
    \(\displaystyle∫\frac{x}{\sqrt{x^2+1}}\,dx\quad\) \(=\quad \displaystyle\sqrt{x^2+1}+C\)

    266) \(\displaystyle∫\frac{x}{\sqrt{1−x^2}}\,dx\)

    267) \(\displaystyle∫(x−1)(x^2−2x)^3\,dx\)

    Answer:
    \(\displaystyle∫(x−1)(x^2−2x)^3\,dx\quad\) \(=\quad\displaystyle\frac{1}{8}(x^2−2x)^4+C\)

    268) \(\displaystyle∫(x^2−2x)(x^3−3x^2)^2\,dx\)

    269) \(\displaystyle\int\cos^3 θ\,dθ\)      (Hint: \(\cos^2 θ=1−\sin^2 θ\))

    Answer:
    \(\displaystyle\int\cos^3 θ\,dθ\quad\) \(=\quad\displaystyle \sin θ−\frac{\sin^3 θ}{3}+C\)

    270) \(\displaystyle\int\sin^3 θ\,dθ\)        (Hint: \(\sin^2 θ=1−\cos^2 θ\))

    271) \(\displaystyle\int x(1−x)^{99}\,dx\)

    Answer:
    \(\displaystyle\int x(1−x)^{99}\,dx\quad\) \(=\quad\displaystyle\frac{(1−x)^{101}}{101}−\frac{(1−x)^{100}}{100}+C\)

    272) \(\displaystyle∫t(1−t^2)^{10}\,dt\)

    273) \(\displaystyle∫(11x−7)^{−3}\,dx\)

    Answer:
    \(\displaystyle−\frac{1}{22(7−11x^2)}+C\)

    274) \(\displaystyle∫(7x−11)^4\,dx\)

    275) \(\displaystyle\int\cos^3 θ\sin θ\,dθ\)

    Answer:
    \(\displaystyle−\frac{\cos^4 θ}{4}+C\)

    281) \(\displaystyle∫\frac{x^2}{(x^3−3)^2}\,dx\)

    Answer:
    \(\displaystyle−\frac{1}{3(x^3−3)}+C\)

    In the following exercises, evaluate the definite integral.

    292) \(\displaystyle∫^1_0x\sqrt{1−x^2}\,dx\)

    293) \(\displaystyle∫^1_0\frac{x}{\sqrt{1+x^2}}\,dx\)

    Answer:
    \(\displaystyle u=1+x^2,du=2x\,dx,\frac{1}{2}∫^2_1u^{−1/2}du=\sqrt{2}−1\)

    294) \(\displaystyle∫^2_0\frac{t}{\sqrt{5+t^2}}\,dt\)

    295) \(\displaystyle∫^1_0\frac{t}{\sqrt{1+t^3}}\,dt\)

    Answer:
    \(\displaystyle u=1+t^3,du=3t^2,\frac{1}{3}∫^2_1u^{−1/2}du=\frac{2}{3}(\sqrt{2}−1)\)

    296) \(\displaystyle\int^{π/4}_0\sec^2 θ\tan θ\,dθ\)

    297) \(\displaystyle\int^{π/4}_0\frac{\sin θ}{\cos^4 θ}\,dθ\)

    Answer:
    \(\displaystyle u=\cos θ,\quad du=−\sin θ\,dθ,\quad∫^1_{1/\sqrt{2}}u^{−4}\,du=\frac{1}{3}(2\sqrt{2}−1)\)

     

    5.6: Integrals Involving Exponential and Logarithmic Functions

    In the following exercises, compute each indefinite integral.

    320) \(\displaystyle ∫e^{2x}\,dx\)

    321) \(\displaystyle ∫e^{−3x}\,dx\)

    Answer:
    \(\displaystyle \frac{−1}{3}e^{−3x}+C\)

    322) \(\displaystyle ∫2^x\,dx\)

    323) \(\displaystyle ∫3^{−x}\,dx\)

    Answer:
    \(\displaystyle −\frac{3^{−x}}{\ln 3}+C\)

    324) \(\displaystyle ∫\frac{1}{2x}\,dx\)

    325) \(\displaystyle ∫\frac{2}{x}\,dx\)

    Answer:
    \(\displaystyle \ln(x^2)+C\)

    326) \(\displaystyle ∫\frac{1}{x^2}\,dx\)

    327) \(\displaystyle ∫\frac{1}{\sqrt{x}}\,dx\)

    Answer:
    \(\displaystyle 2\sqrt{x}+C\)

     

    In the following exercises, find each indefinite integral by using appropriate substitutions.

    328) \(\displaystyle ∫\frac{\ln x}{x}\,dx\)

    329) \(\displaystyle ∫\frac{\,dx}{x(\ln x)^2}\)

    Answer:
    \(\displaystyle −\frac{1}{\ln x}+C\)

     

    336) \(\displaystyle ∫xe^{−x^2}\,dx\)

    337) \(\displaystyle ∫x^2e^{−x^3}\,dx\)

    Answer:
    \(\displaystyle \frac{−e^{−x^3}}{3}+C\)

    338) \(\displaystyle ∫e^{\sin x}\cos x\,dx\)

    339) \(\displaystyle ∫e^{\tan x}\sec^2 x\,dx\)

    Answer:
    \(\displaystyle e^{\tan x}+C\)

    340) \(\displaystyle ∫\frac{e^{\ln x}}{x}\,dx\)

    341) \(\displaystyle ∫\frac{e^{\ln(1−t)}}{1−t}\,dt\)

    Answer:
    \(\displaystyle t+C\)

     

    In the following exercises, evaluate the definite integral.

    355) \(\displaystyle ∫^2_1\frac{1+2x+x^2}{3x+3x^2+x^3}\,dx\)

    Answer:
    \(\displaystyle \frac{1}{3}\ln(\frac{26}{7})\)

    356) \(\displaystyle ∫^{π/4}_0\tan x\,dx\)

    357) \(\displaystyle ∫^{π/3}_0\frac{\sin x−\cos x}{\sin x+\cos x}\,dx\)

    Answer:
    \(\displaystyle ln(\sqrt{3}−1)\)

    358) \(\displaystyle ∫^{π/2}_{π/6}\csc x\,dx\)

    359) \(\displaystyle ∫^{π/3}_{π/4}\cot x\,dx\)

    Answer:
    \(\displaystyle \frac{1}{2}\ln\frac{3}{2}\)

     

    In the following exercises, integrate using the indicated substitution.

    360) \(\displaystyle ∫\frac{x}{x−100}\,dx;\quad u=x−100\)

    361) \(\displaystyle ∫\frac{y−1}{y+1}dy;\quad u=y+1\)

    Answer:
    \(\displaystyle y−2\ln|y+1|+C\)

    362) \(\displaystyle ∫\frac{1−x^2}{3x−x^3}\,dx;\quad u=3x−x^3\)

    363) \(\displaystyle ∫\frac{\sin x+\cos x}{\sin x−\cos x}\,dx;\quad u=\sin x−\cos x\)

    Answer:
    \(\displaystyle \ln|\sin x−\cos x|+C\)

    364) \(\displaystyle ∫e^{2x}\sqrt{1−e^{2x}}\,dx;\quad u=e^{2x}\)

    365) \(\displaystyle ∫\ln(x)\frac{\sqrt{1−(\ln x)^2}}{x}\,dx;\quad u=\ln x\)

    Answer:
    \(\displaystyle −\frac{1}{3}(1−(\ln x^2))^{3/2}+C\)

     

     

     

    In the following exercises, \(\displaystyle f(x)≥0\) for \(\displaystyle a≤x≤b\). Find the area under the graph of \(\displaystyle f(x)\) between the given values \(a\) and \(b\) by integrating.

    372) \(\displaystyle f(x)=\frac{\log_{10}(x)}{x};\quad a=10,b=100\)

    373) \(\displaystyle f(x)=\frac{\log_2(x)}{x};\quad a=32,b=64\)

    Answer:
    \(\displaystyle \frac{11}{2}\ln 2\)

    374) \(\displaystyle f(x)=2^{−x};\quad a=1,b=2\)

    375) \(\displaystyle f(x)=2^{−x};\quad a=3,b=4\)

    Answer:
    \(\displaystyle \frac{1}{\ln(65,536)}\)

    376) Find the area under the graph of the function \(\displaystyle f(x)=xe^{−x^2}\) between \(\displaystyle x=0\) and \(\displaystyle x=5\).

    377) Compute the integral of \(\displaystyle f(x)=xe^{−x^2}\) and find the smallest value of N such that the area under the graph \(\displaystyle f(x)=xe^{−x^2}\) between \(\displaystyle x=N\) and \(\displaystyle x=N+10\) is, at most, 0.01.

    Answer:
    \(\displaystyle ∫^{N+1}_Nxe^{−x^2}\,dx=\frac{1}{2}(e^{−N^2}−e^{−(N+1)^2}).\) The quantity is less than 0.01 when \(\displaystyle N=2\).

    378) Find the limit, as N tends to infinity, of the area under the graph of \(\displaystyle f(x)=xe^{−x^2}\) between \(\displaystyle x=0\) and \(\displaystyle x=5\).

    379) Show that \(\displaystyle ∫^b_a\frac{\,dt}{t}=∫^{1/a}_{1/b}\frac{\,dt}{t}\) when \(\displaystyle 0<a≤b\).

    Answer:
    \(\displaystyle ∫^b_a\frac{\,dx}{x}=\ln(b)−\ln(a)=\ln(\frac{1}{a})−\ln(\frac{1}{b})=∫^{1/a}_{1/b}\frac{\,dx}{x}\)

    380) Suppose that \(\displaystyle f(x)>0\) for all x and that f and g are differentiable. Use the identity \(\displaystyle f^g=e^{glnf}\) and the chain rule to find the derivative of \(\displaystyle f^g\).

    381) Use the previous exercise to find the antiderivative of \(\displaystyle h(x)=x^x(1+\ln x)\) and evaluate \(\displaystyle ∫^3_2x^x(1+\ln x)\,dx\).

    Answer:
    23

    382) Show that if \(\displaystyle c>0\), then the integral of \(\displaystyle 1/x\) from ac to bc \(\displaystyle (0<a<b)\) is the same as the integral of \(\displaystyle 1/x\) from a to b.

     

    The following exercises are intended to derive the fundamental properties of the natural log starting from the definition \(\displaystyle \ln(x)=∫^x_1\frac{\,dt}{t}\), using properties of the definite integral and making no further assumptions.

    383) Use the identity \(\displaystyle \ln(x)=∫^x_1\frac{\,dt}{t}\) to derive the identity \(\displaystyle \ln(\frac{1}{x})=−\ln x\).

    Solution: We may assume that \(\displaystyle x>1\),so \(\displaystyle \frac{1}{x}<1.\) Then, \(\displaystyle ∫^{1/x}_{1}\frac{\,dt}{t}\). Now make the substitution \(\displaystyle u=\frac{1}{t}\), so \(\displaystyle du=−\frac{\,dt}{t^2}\) and \(\displaystyle \frac{du}{u}=−\frac{\,dt}{t}\), and change endpoints: \(\displaystyle ∫^{1/x}_1\frac{\,dt}{t}=−∫^x_1\frac{du}{u}=−lnx.\)

    384) Use a change of variable in the integral \(\displaystyle ∫^{xy}_1\frac{1}{t}\,dt\) to show that \(\displaystyle lnxy=lnx+lny\) for \(\displaystyle x,y>0\).

    385) Use the identity \(\displaystyle \ln x=∫^x_1\frac{\,dt}{x}\) to show that \(\displaystyle \ln(x)\) is an increasing function of \(x\) on \(\displaystyle [0,∞)\), and use the previous exercises to show that the range of \(\displaystyle \ln(x)\) is \(\displaystyle (−∞,∞)\). Without any further assumptions, conclude that \(\displaystyle \ln(x)\) has an inverse function defined on \(\displaystyle (−∞,∞).\)

    386) Pretend, for the moment, that we do not know that \(\displaystyle e^x\) is the inverse function of \(\displaystyle \ln(x)\), but keep in mind that \(\displaystyle \ln(x)\) has an inverse function defined on \(\displaystyle (−∞,∞)\). Call it \(E\). Use the identity \(\displaystyle \ln xy=\ln x+\ln y\) to deduce that \(\displaystyle E(a+b)=E(a)E(b)\) for any real numbers \(a\), \(b\).

    387) Pretend, for the moment, that we do not know that \(\displaystyle e^x\) is the inverse function of \(\displaystyle \ln x\), but keep in mind that \(\displaystyle \ln x\) has an inverse function defined on \(\displaystyle (−∞,∞)\). Call it \(E\). Show that \(\displaystyle E'(t)=E(t).\)

    Solution: \(\displaystyle x=E(\ln(x)).\) Then, \(\displaystyle 1=\frac{E'(\ln x)}{x}\) or \(\displaystyle x=E'(\ln x)\). Since any number t can be written \(\displaystyle t=\ln x\) for some \(x\), and for such \(t\) we have \(\displaystyle x=E(t)\), it follows that for any \(\displaystyle t,\,E'(t)=E(t).\)

    388) The sine integral, defined as \(\displaystyle S(x)=∫^x_0\frac{\sin t}{t}\,dt\) is an important quantity in engineering. Although it does not have a simple closed formula, it is possible to estimate its behavior for large \(x\). Show that for \(\displaystyle k≥1,|S(2πk)−S(2π(k+1))|≤\frac{1}{k(2k+1)π}.\) (Hint: \(\displaystyle \sin(t+π)=−\sin t\))

    389) [T] The normal distribution in probability is given by \(\displaystyle p(x)=\frac{1}{σ\sqrt{2π}}e^{−(x−μ)^2/2σ^2}\), where \(σ\) is the standard deviation and \(μ\) is the average. The standard normal distribution in probability, \(\displaystyle p_s\), corresponds to \(\displaystyle μ=0\) and \(\displaystyle σ=1\). Compute the left endpoint estimates \(\displaystyle R_{10}\) and \(\displaystyle R_{100}\) of \(\displaystyle ∫^1_{−1}\frac{1}{\sqrt{2π}}e^{−x^{2/2}}\,dx.\)

    Solution: \(\displaystyle R_{10}=0.6811,R_{100}=0.6827\)

    390) [T] Compute the right endpoint estimates \(\displaystyle R_{50}\) and \(\displaystyle R_{100}\) of \(\displaystyle ∫^5_{−3}\frac{1}{2\sqrt{2π}}e^{−(x−1)^2/8}\).