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Mathematics LibreTexts

4.7 E Exercises

  • Page ID
    13774
  • [ "stage:draft", "article:topic" ]

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    Exercise \(\PageIndex{1}\)

    In the following exercises, use a change of variables to evaluate the definite integral.

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

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

    Answer

    \(\displaystyleu=1+x^2,du=2xdx,\frac{1}{2}∫^2_1u^{−1/2}du=\sqrt{2}−1\)

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

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

    Answer:

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

     

    5) \(\displaystyle∫^{π/4}_0sec^2θtanθdθ\)

    6) \(\displaystyle∫^{π/4}_0\frac{sinθ}{cos^4θ}dθ\)

    Answer:

    \(\displaystyleu=cosθ,du=−sinθdθ,∫^1_{1/\sqrt{2}}u^{−4}du=\frac{1}{3}(2\sqrt{2}−1)\)

     

    Exercise \(\PageIndex{2}\)

    In the following exercises, evaluate the definite integral.

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

    Answer

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

    2) \(\displaystyle ∫^{π/4}_0tanxdx\)

    3) \(\displaystyle ∫^{π/3}_0\frac{sinx−cosx}{sinx+cosx}dx\)

    Answer:

    \(\displaystyle ln(\sqrt{3}−1)\)

     4)  \(\displaystyle ∫^{π/2}_{π/6}cscxdx\)

    5) \(\displaystyle ∫^{π/3}_{π/4}cotxdx\)

    Answer:

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

    Exercise \(\PageIndex{3}\)

    In the following exercises, use a change of variables to show that each definite integral is equal to zero.

    1. \(\displaystyle∫^π_0cos^2(2θ)sin(2θ)dθ\)
    2.  \(\displaystyle∫^\sqrt{π}_0tcos(t^2)sin(t^2)dt\)

    Answer:

    \(\displaystyle u=sin(t^2);\) the integral becomes \(\displaystyle\frac{1}{2}∫^0_0udu.\)

     

     3.  \(\displaystyle∫^1_0(1−2t)dt\)

    4.  \(\displaystyle∫^1_0\frac{1−2t}{(1+(t−\frac{1}{2})^2)}dt\)

    Answer:

    \(\displaystyleu=(1+(t−\frac{1}{2})^2);\) the integral becomes \(\displaystyle−∫^{5/4}_{5/4}\frac{1}{u}du\).

    5. \(\displaystyle∫^π_0sin((t−\frac{π}{2})^3)cos(t−\frac{π}{2})dt\)

    6. \(\displaystyle∫^2_0(1−t)cos(πt)dt\)

    Answer

     

    \(\displaystyle∫^{−1}_1ucos(π(1−u))du=∫^{−1}_1u[cosπcosu−sinπsinu]du=−∫^{−1}_1ucosudu=∫^{1−}_1ucosudu=0\)

    since the integrand is odd.

    7.  \(\displaystyle∫^{3π/4}_{π/4}sin^2tcostdt\)

    Exercise \(\PageIndex{4}\)

    In the following exercises, evaluate the definite integral.

    Answer

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