Calculus II - Trig Substitutions (Assignment Problems)

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Section 7.3 : Trig Substitutions

For problems 1 – 15 use a trig substitution to eliminate the root.

$\sqrt {64{t^2} + 1}$
$\sqrt {4{z^2} - 49}$
$\sqrt {7 - {w^2}}$
${\left( {16 - 81{x^2}} \right)^{\frac{7}{2}}}$
$\sqrt {6 + 9{y^2}}$
${\left( {1 - 8{z^2}} \right)^{\frac{3}{2}}}$
$\sqrt {9 - 16{{\left( {3x - 1} \right)}^2}}$
${\left( {11 + {{\left( {{t^2} + 1} \right)}^2}} \right)^{\frac{5}{2}}}$
$\sqrt {144{{\left( {z + 8} \right)}^2} - 3}$
$\sqrt {4{x^2} - 24x + 43}$
${\left( {2{z^2} - 24z + 36} \right)^{\frac{{11}}{2}}}$
$\sqrt { - 4 - 10t - 5{t^2}}$
$\sqrt {9{{\sin }^2}\left( {4t} \right) - 1}$
$\sqrt {36 - 9{{\bf{e}}^{3z}}}$
$\sqrt {x + 16}$

For problems 16 – 32 use a trig substitution to evaluate the given integral.

$\displaystyle \int{{3{x^5}\sqrt {16 - {x^2}} \,dx}}$
$\displaystyle \int{{{t^3}{{\left( {25 + 81{t^2}} \right)}^{\frac{5}{2}}}\,dt}}$
$\displaystyle \int_{0}^{{\frac{1}{4}}}{{\frac{{{w^3}}}{{\sqrt {1 - 9{w^2}} }}\,dw}}$
$\displaystyle \int{{\frac{{{z^5}}}{{{{\left( {9{z^2} - 25} \right)}^{\frac{3}{2}}}}}\,dz}}$
$\displaystyle \int_{{ - 3}}^{{ - 1}}{{{y^3}\sqrt {49{y^2} - 4} \,dy}}$
$\displaystyle \int_{1}^{5}{{\frac{5}{{{x^2}\sqrt {{x^2} + 4} }}\,dx}}$
$\displaystyle \int{{\frac{{\sqrt {3 - 4{t^2}} }}{{{t^2}}}\,dt}}$
$\displaystyle \int{{\frac{{{w^5}}}{{\sqrt {8{w^2} + 1} }}\,dw}}$
$\displaystyle \int{{\frac{{\sqrt {{x^2} - 15} }}{{{x^3}}}\,dx}}$
$\displaystyle \int{{\frac{2}{{{{\left( {x - 3} \right)}^6}\sqrt { - {x^2} + 6x - 5} }}\,dx}}$
$\displaystyle \int{{\frac{1}{{{{\left( {z + 1} \right)}^2}{{\left( {2{z^2} + 4z - 34} \right)}^{\frac{3}{2}}}}}\,dz}}$
$\displaystyle \int{{\frac{{\sqrt {4{y^2} - 16y + 19} }}{{{{\left( {y - 2} \right)}^6}}}\,dy}}$
$\displaystyle \int_{9}^{{12}}{{\frac{{{{\left( {t - 4} \right)}^3}}}{{\sqrt {{t^2} - 8t + 7} }}\,dt}}$
$\displaystyle \int_{0}^{6}{{\sqrt {5{x^2} + 10x + 6} \,dx}}$
$\displaystyle \int{{{x^7}\sqrt {9 - {x^4}} \,dx}}$
$\displaystyle \int{{\frac{{{{\bf{e}}^{12t}}}}{{\sqrt {4{{\bf{e}}^{6t}} - 1} }}\,dt}}$
$\displaystyle \int{{\sin \left( z \right){{\cos }^3}\left( z \right)\sqrt {16 + {{\cos }^2}\left( z \right)} \,dz}}$