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Mathematics LibreTexts

7.3E: Exercises for Trigonometric Substitution

  • Page ID
    18289
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    Simplify the expressions in exercises 1 - 5 by writing each one using a single trigonometric function.

    1) \(4−4\sin^2θ\)

    2) \(9\sec^2θ−9\)

    Answer:
    \(9\sec^2θ−9 \quad = \quad 9\tan^2θ\)

    3) \(a^2+a^2\tan^2θ\)

    4) \(a^2+a^2\sinh^2θ\)

    Answer:
    \(a^2+a^2\sinh^2θ \quad = \quad a^2\cosh^2θ\)

    5) \(16\cosh^2θ−16\)

    Use the technique of completing the square to express each trinomial in exercises 6 - 8 as the square of a binomial.

    6) \(4x^2−4x+1\)

    Answer:
    \( 4(x−\frac{1}{2})^2\)

    7) \(2x^2−8x+3\)

    8) \(−x^2−2x+4\)

    Answer:
    \( −(x+1)^2+5\)

    In exercises 9 - 28, integrate using the method of trigonometric substitution. Express the final answer in terms of the original variable.

    9) \(\displaystyle ∫\frac{dx}{\sqrt{4−x^2}}\)

    10) \(\displaystyle ∫\frac{dx}{\sqrt{x^2−a^2}}\)

    Answer:
    \(\displaystyle ∫\frac{dx}{\sqrt{x^2−a^2}} \quad = \quad \ln∣x+\sqrt{−a^2+x^2}∣+C\)

    11) \(\displaystyle ∫\sqrt{4−x^2}\,dx\)

    12) \(\displaystyle ∫\frac{dx}{\sqrt{1+9x^2}}\)

    Answer:
    \(\displaystyle ∫\frac{dx}{\sqrt{1+9x^2}} \quad = \quad \tfrac{1}{3}\ln∣\sqrt{9x^2+1}+3x∣+C\)

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

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

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

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

    16) \(\displaystyle ∫\sqrt{x^2+9}\,dx\)

    Answer:
    \(\displaystyle ∫\sqrt{x^2+9}\,dx \quad = \quad 9\left[\frac{x\sqrt{x^2+9}}{18}+\tfrac{1}{2}\ln\left|\frac{\sqrt{x^2+9}}{3}+\frac{x}{3}\right|\right]+C\)

    17) \(\displaystyle ∫\frac{\sqrt{x^2−25}}{x}\,dx\)

    18) \(\displaystyle ∫\frac{θ^3}{\sqrt{9−θ^2}}\,dθ\)

    Answer:
    \(\displaystyle ∫\frac{θ^3dθ}{\sqrt{9−θ^2}}\,dθ \quad = \quad −\tfrac{1}{3}\sqrt{9−θ^2}(18+θ^2)+C\)

    19) \(\displaystyle ∫\frac{dx}{\sqrt{x^6−x^2}}\)

    20) \(\displaystyle ∫\sqrt{x^6−x^8}\,dx\)

    Answer:
    \(\displaystyle ∫\sqrt{x^6−x^8}\,dx \quad = \quad \frac{(−1+x^2)(2+3x^2)\sqrt{x^6−x^8}}{15x^3}+C\)

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

    22) \(\displaystyle ∫\frac{dx}{(x^2−9)^{3/2}}\)

    Answer:
    \(\displaystyle ∫\frac{dx}{(x^2−9)^{3/2}} \quad = \quad −\frac{x}{9−\sqrt{9+x^2}}+C\)

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

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

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

    25) \(\displaystyle ∫\frac{x^2}{x^2+4}\,dx\)

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

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

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

    28) \(\displaystyle ∫^1_{−1}(1−x^2)^{3/2}\,dx\)

    Answer:
    \(\displaystyle ∫^1_{−1}(1−x^2)^{3/2}\,dx \quad = \quad \tfrac{1}{8}\left(x(5−2x^2)\sqrt{1−x^2}+3\arcsin x\right)+C\)

    In exercises 29 - 34, use the substitutions \(x=\sinh θ, \, \cosh θ,\) or \(\tanh θ.\) Express the final answers in terms of the variable \(x\).

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

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

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

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

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

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

    33) \(\displaystyle ∫\frac{dx}{1−x^2}\)

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

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

    Use the technique of completing the square to evaluate the integrals in exercises 35 - 39.

    35) \(\displaystyle ∫\frac{1}{x^2−6x}\,dx\)

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

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

    37) \(\displaystyle ∫\frac{1}{\sqrt{−x^2+2x+8}}\,dx\)

    38) \(\displaystyle ∫\frac{1}{\sqrt{−x^2+10x}}\,dx\)

    Answer:
    \(\displaystyle ∫\frac{1}{\sqrt{−x^2+10x}}\,dx \quad = \quad \frac{2\sqrt{−10+x}\sqrt{x}\ln\left|\sqrt{−10+x}+\sqrt{x}\right|}{\sqrt{(10−x)x}}+C\)

    39) \(\displaystyle ∫\frac{1}{\sqrt{x^2+4x−12}}\,dx\)

    40) Evaluate the integral without using calculus: \(\displaystyle ∫^3_{−3}\sqrt{9−x^2}\,dx.\)

    Answer:
    \(\displaystyle ∫^3_{−3}\sqrt{9−x^2}\,dx \quad = \quad \frac{9π}{2}\); area of a semicircle with radius 3

    41) Find the area enclosed by the ellipse \(\dfrac{x^2}{4}+\dfrac{y^2}{9}=1.\)

    42) Evaluate the integral \(\displaystyle ∫\frac{dx}{\sqrt{1−x^2}}\) using two different substitutions. First, let \(x=\cos θ\) and evaluate using trigonometric substitution. Second, let \(x=\sin θ\) and use trigonometric substitution. Are the answers the same?

    Answer:
    \(\displaystyle ∫\frac{dx}{\sqrt{1−x^2}} \quad = \quad \arcsin(x)+C\) is the common answer.

    43) Evaluate the integral \(\displaystyle ∫\frac{dx}{x\sqrt{x^2−1}}\) using the substitution \(x=\sec θ\). Next, evaluate the same integral using the substitution \(x=\csc θ.\) Show that the results are equivalent.

    44) Evaluate the integral \(\displaystyle ∫\frac{x}{x^2+1}\,dx\) using the form \(\displaystyle ∫\frac{1}{u}\,du\). Next, evaluate the same integral using \(x=\tan θ.\) Are the results the same?

    Answer:
    \(\displaystyle ∫\frac{x}{x^2+1}\,dx \quad = \quad \frac{1}{2}\ln(1+x^2)+C\) is the result using either method.

    45) State the method of integration you would use to evaluate the integral \(\displaystyle ∫x\sqrt{x^2+1}\,dx.\) Why did you choose this method?

    46) State the method of integration you would use to evaluate the integral \(\displaystyle ∫x^2\sqrt{x^2−1}\,dx.\) Why did you choose this method?

    Answer:
    Use trigonometric substitution. Let \(x=\sec(θ).\)

    47) Evaluate \(\displaystyle ∫^1_{−1}\frac{x}{x^2+1}\,dx\)

    48) Find the length of the arc of the curve over the specified interval: \(y=\ln x,\quad [1,5].\) Round the answer to three decimal places.

    Answer:
    \( s = 4.367\) units

    49) Find the surface area of the solid generated by revolving the region bounded by the graphs of \(y=x^2,\, y=0,\, x=0\), and \(x=\sqrt{2}\) about the \(x\)-axis. (Round the answer to three decimal places).

    50) The region bounded by the graph of \(f(x)=\dfrac{1}{1+x^2}\) and the \(x\)-axis between \(x=0\) and \(x=1\) is revolved about the \(x\)-axis. Find the volume of the solid that is generated.

    Answer:
    \( V = \left(\frac{π^2}{8}+\frac{π}{4}\right) \, \text{units}^3\)

    In exercises 51 - 52, solve the initial-value problem for \(y\) as a function of \(x\).

    51) \((x^2+36)\dfrac{dy}{dx}=1, \quad y(6)=0\)

    52) \((64−x^2)\dfrac{dy}{dx}=1, \quad y(0)=3\)

    Answer:
    \( y=\tfrac{1}{16}\ln\left|\dfrac{x+8}{x−8}\right|+3\)

    53) Find the area bounded by \(y=\dfrac{2}{\sqrt{64−4x^2}},\, x=0,\, y=0\), and \(x=2\).

    54) An oil storage tank can be described as the volume generated by revolving the area bounded by \(y=\dfrac{16}{\sqrt{64+x^2}},\, x=0,\, y=0,\, x=2\) about the \(x\)-axis. Find the volume of the tank (in cubic meters).

    Answer:
    \(V = 24.6\) m3

    55) During each cycle, the velocity \(v\) (in feet per second) of a robotic welding device is given by \(v=2t−\dfrac{14}{4+t^2}\), where \(t\) is time in seconds. Find the expression for the displacement \(s\) (in feet) as a function of \(t\) if \(s=0\) when \(t=0\).

    56) Find the length of the curve \(y=\sqrt{16−x^2}\) between \(x=0\) and \(x=2\).

    Answer:
    \( s = \frac{2π}{3}\) units

    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.