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Mathematics LibreTexts

4.1 E Exercises

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  • Page ID
    13730
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    Exercise \(\PageIndex{1}\)

    1. Why is u-substitution referred to as change of variable?

    2. If \(\displaystyle f=g∘h\), when reversing the chain rule, \(\displaystyle \frac{d}{d}x(g∘h)(x)=g′(h(x))h′(x)\), should you take \(\displaystyle u=g(x)\) or u= \(\displaystyle h(x)?\)

    Answer

    \(\displaystyle u=h(x)\).

     
    Exercise \(\PageIndex{2}\)

    In the following exercises, verify each identity using differentiation. Then, using the indicated u-substitution, identify f such that the integral takes the form \(\displaystyle∫f(u)du.\)

    1. \(\displaystyle∫x\sqrt{x+1}x=\frac{2}{15}(x+1)^{3/2}(3x−2)+C;u=x+1\)

    2. \(\displaystyle∫\frac{x^2}{\sqrt{x−1}}dx(x>1)=\frac{2}{15}\sqrt{x−1}(3x^2+4x+8)+C;u=x−1\)

    Answer

    \(\displaystyle f(u)=\frac{(u+1)^2}{\sqrt{u}}\)

    3. \(\displaystyle∫x\sqrt{4x^2+9}dx=\frac{1}{12}(4x^2+9)^{3/2}+C;u=4x^2+9\)

    4. \(\displaystyle∫\frac{x}{\sqrt{4x^2+9}}dx=\frac{1}{4}\sqrt{4x^2+9}+C;u=4x^2+9\)

    Answer

    \(\displaystyle du=8xdx;f(u)=\frac{1}{8\sqrt{u}}\)

    5. \(\displaystyle∫\frac{x}{(4x^2+9)^2}dx=−\frac{1}{8(4x^2+9)};u=4x^2+9\)

    Answer

    \(\displaystyle du=8xdx;f(u)=\frac{1}{8u^2}\)

     

    Exercise \(\PageIndex{3}\)

    In the following exercises, find the antiderivative using the indicated substitution.

    1. \(\displaystyle ∫(x+1)^4dx;u=x+1\)

    Answer

    \(\displaystyle \frac{1}{5}(x+1)^5+C\)

    2. \(\displaystyle∫(x−1)^5dx;u=x−1\)

    3. \(\displaystyle∫(2x−3)^{−7}dx;u=2x−3\)

    Answer

    \(\displaystyle−\frac{1}{12(3−2x)^6}+C\)

    4. \(\displaystyle∫(3x−2)^{−11}dx;u=3x−2\)

    5. \(\displaystyle∫\frac{x}{\sqrt{x^2+1}}dx;u=x^2+1\)

    Answer

    \(\displaystyle\sqrt{x^2+1}+C\)

    6. \(\displaystyle∫\frac{x}{\sqrt{1−x^2}}dx;u=1−x^2\)

    7. \(\displaystyle∫(x−1)(x^2−2x)^3dx;u=x^2−2x\)

    Answer

    \(\displaystyle\frac{1}{8}(x^2−2x)^4+C\)

    8. \(\displaystyle∫(x^2−2x)(x^3−3x^2)^2dx;u=x^3=3x^2\)

    9. \(\displaystyle∫cos^3θdθ;u=sinθ (Hint:cos^2θ=1−sin^2θ)\)

    Answer

    \(\displaystylesinθ−\frac{sin^3θ}{3}+C\)

    10. \(\displaystyle∫sin^3θdθ;u=cosθ (Hint:sin^2θ=1−cos^2θ)\)

     

    Exercise \(\PageIndex{4}\)

    In the following exercises, use a suitable change of variables to determine the indefinite integral.

    1. \(\displaystyle∫x(1−x)^{99}dx\)

    Answer

     \(\displaystyle\frac{(1−x)^{101}}{101}−\frac{(1−x)^{100}}{100}+C\)

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

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

    Answer

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

    4. \displaystyle∫(7x−11)^4dx\)

    5. \(\displaystyle∫cos^3θsinθdθ\)

    Answer

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

    6. \(\displaystyle∫sin^7θcosθdθ\)

    7. \(\displaystyle∫cos^2(πt)sin(πt)dt\)

    Answer

    \(\displaystyle−\frac{cos^3(πt)}{3π}+C\)

    8. \(\displaystyle∫sin^2xcos^3xdx (Hint:sin^2x+cos^2x=1)\)

    9. \(\displaystyle∫tsin(t^2)cos(t^2)dt\)

    Answer

    \(\displaystyle−\frac{1}{4}cos^2(t^2)+C\)

    10. \(\displaystyle∫t^2cos^2(t^3)sin(t^3)dt\)

    11. \(\displaystyle∫\frac{x^2}{(x^3−3)^2}dx\)

    Answer

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

    12.  \(\displaystyle∫\frac{x^3}{\sqrt{1−x^2}}dx\)

    13. \(\displaystyle∫\frac{y^5}{(1−y^3)^{3/2}}dy\)

    Answer

    \(\displaystyle−\frac{2(y^3−2)}{3\sqrt{1−y^3}}\)

    14. \(\displaystyle∫cosθ(1−cosθ)^{99}sinθdθ\)

    15.  \(\displaystyle∫(1−cos^3θ)^{10}cos^2θsinθdθ\)

    Answer

    \(\displaystyle\frac{1}{33}(1−cos^3θ)^{11}+C\)

    16. \(\displaystyle∫(cosθ−1)(cos^2θ−2cosθ)^3sinθdθ\)

    17. \(\displaystyle∫(sin^2θ−2sinθ)(sin^3θ−3sin^2θ^)3cosθdθ\)

    Answer

    \(\displaystyle\frac{1}{12}(sin^3θ−3sin^2θ)^4+C\)

     

    Exercise \(\PageIndex{5}\)

    1) \(\displaystyle \int \frac{ x^3\ dx}{\sqrt{1-x^2}} \)

    2) \(\displaystyle\int (\csc x)(\cot x)(e^{\csc x})\,\,dx\)

    3)\(\displaystyle \int \frac{1}{e^x+e^{-x}}dx\)

    4) \(\displaystyle \int \frac {1}{\sqrt{4-x^2}} \ dx \)

    5) \(\displaystyle \int \frac{\sqrt{x^2-4}}{x} \ dx \)

    6) \(\displaystyle \int \frac{1}{x^2 + 6x + 13} dx\)

    Answer

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    Contributors

    • Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

    • Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada)