Exercises: Trigonometric Substitions

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Various exercises relating to trigonometric substitutions.

Compute the indefinite integrals below. Since there are many possible answers (which differ by constant values), use the given instructions if needed to choose which possible answer to use. Do not forget absolute value signs inside logarithms when they are needed.

$\int x^2\sqrt {1-x^2}\ dx = \answer {\frac {1}{8}\arcsin x-\frac {1}{8}x\sqrt {1-x^2}(1-2x^2)}+C$ (Choose your answer to equal $0$ at $x = 0$ .)

$\int \frac {1}{(x^2+1)^2}\ dx = \answer {\frac {1}{2}\left (\arctan x+\frac {x}{x^2+1}\right )} +C$ (Choose your answer to equal $0$ at $x = 0$ .)

$\int \frac {x^2}{\sqrt {x^2+3}}\ dx = \answer {\frac {1}{2}x\sqrt {x^2+3}-\frac {3}{2}\ln \left |\frac {\sqrt {x^2+3}}{\sqrt {3}}+\frac {x}{\sqrt {3}}\right |} +C$ (Choose your answer to equal $0$ at $x = 0$ .)

Sample Quiz Questions

Compute the integral $\int _{-2}^{2}\frac {5}{(5-x^2)^{3/2}}~dx.$ (Hints won’t be revealed until after you choose a response.)

$\displaystyle 2$ $\displaystyle 3$ $\displaystyle 4$ $\displaystyle 5$ $\displaystyle 6$ $\displaystyle 7$

Begin by making the trig substitution $x=\sqrt {5}\sin \theta$ .

It follows that $\begin {aligned} \int \frac {5}{(5-x^2)^{3/2}}~dx & = \int \frac {5}{(5-(\sqrt {5}\sin \theta )^2)^{3/2}} \cdot (\sqrt {5}\cos \theta )~d \theta \\ & = \int (\cos \theta )^{-2}~d \theta \\ & = \int (\sec \theta )^{2} ~ d \theta = (\tan \theta ) + C. \end {aligned}$

To finish, use the inversion identity $\tan \theta = \frac {x}{\sqrt {5-x^2}}.$ Therefore $\int _{-2}^{2}\frac {5}{(5-x^2)^{3/2}}~dx = \left .\frac {x}{\sqrt {5-x^2}}\right |_{-2}^{2} = \left (2\right ) - \left (-2\right ) = 4.$

Compute the integral $\int _{-1}^{1}\frac {3}{(3+x^2)^{3/2}}~dx.$ (Hints won’t be revealed until after you choose a response.)

$\displaystyle \frac {1}{2}$ $\displaystyle 1$ $\displaystyle \frac {3}{2}$ $\displaystyle 2$ $\displaystyle \frac {5}{2}$ $\displaystyle 3$

Begin by making the trig substitution $x=\sqrt {3}\tan \theta$ .

It follows that $\begin {aligned} \int \frac {3}{(3+x^2)^{3/2}}~dx & = \int \frac {3}{(3+(\sqrt {3}\tan \theta )^2)^{3/2}} \cdot (\sqrt {3}\sec ^2 \theta )~d \theta \\ & = \int (\sec \theta )^{-1}~d \theta \\ & = \int (\cos \theta ) ~ d \theta = (\sin \theta ) + C. \end {aligned}$

To finish, use the inversion identity $\cos \theta = \frac {x}{\sqrt {3+x^2}}.$ Therefore $\int _{-1}^{1}\frac {3}{(3+x^2)^{3/2}}~dx = \left .\frac {x}{\sqrt {3+x^2}}\right |_{-1}^{1} = \left (\frac {1}{2}\right ) - \left (-\frac {1}{2}\right ) = 1.$

Compute the integral $\int _{4}^{5}\frac {16\sqrt {x^2-16}}{x^{4}}~dx.$ (Hints won’t be revealed until after you choose a response.)

$\displaystyle \frac {4}{125}$ $\displaystyle \frac {1}{25}$ $\displaystyle \frac {6}{125}$ $\displaystyle \frac {7}{125}$ $\displaystyle \frac {8}{125}$ $\displaystyle \frac {9}{125}$

Begin by making the trig substitution $x=4\sec \theta$ .

It follows that $\begin {aligned} \int \frac {16\sqrt {x^2-16}}{x^{4}}~dx & = \int \frac {16\sqrt {(4\sec \theta )^2-16}}{(4\sec \theta )^{4}} \cdot (4\sec \theta \tan \theta )~d \theta \\ & = \int (\sec \theta )^{-3}(\tan \theta )^{2}~d \theta \\ & = \int (\sin \theta )^{2}(\cos \theta ) ~ d \theta = \frac {1}{3}(\sin \theta )^{3} + C. \end {aligned}$

To finish, use the inversion identity $\sin \theta = \frac {\sqrt {x^2-16}}{x}.$ Therefore $\int _{4}^{5}\frac {16\sqrt {x^2-16}}{x^{4}}~dx = \left .\frac {1}{3}\frac {(x^2-16)^{3/2}}{x^{3}}\right |_{4}^{5} = \left (\frac {9}{125}\right ) - \left (0\right ) = \frac {9}{125}.$

Sample Exam Questions

Compute the value of the integral below. $\int _0^{\frac {1}{\sqrt {2}}} \frac {1}{(1-x^2)^{\frac {3}{2}}} dx$

$0$ $1$ $2$ $3$ $4$ none of these

Evaluate $\displaystyle \int _0^3 \frac {dx}{(25-x^2)^{3/2}}$ .

$0$ $\displaystyle \frac {1}{100}$ $\displaystyle \frac {3}{100}$ $\displaystyle \frac {5}{100}$ $\displaystyle \frac {7}{100}$ none of these