Calculus II - Trig Substitutions (Practice Problems)

Hide Mobile Notice

Mobile Notice

You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 7.3 : Trig Substitutions

For problems 1 – 8 use a trig substitution to eliminate the root.

$\sqrt {4 - 9{z^2}}$ Solution
$\sqrt {13 + 25{x^2}}$ Solution
${\left( {7{t^2} - 3} \right)^{\frac{5}{2}}}$ Solution
$\sqrt {{{\left( {w + 3} \right)}^2} - 100}$ Solution
$\sqrt {4{{\left( {9t - 5} \right)}^2} + 1}$ Solution
$\sqrt {1 - 4z - 2{z^2}}$ Solution
${\left( {{x^2} - 8x + 21} \right)^{\frac{3}{2}}}$ Solution
$\sqrt {{{\bf{e}}^{8x}} - 9}$ Solution

For problems 9 – 16 use a trig substitution to evaluate the given integral.

$\displaystyle \int{{\frac{{\sqrt {{x^2} + 16} }}{{{x^4}}}\,dx}}$ Solution
$\displaystyle \int{{\sqrt {1 - 7{w^2}} \,dw}}$ Solution
$\displaystyle \int{{{t^3}{{\left( {3{t^2} - 4} \right)}^{\frac{5}{2}}}\,dt}}$ Solution
$\displaystyle \int_{{ - 7}}^{{ - 5}}{{\frac{2}{{{y^4}\sqrt {{y^2} - 25} }}\,dy}}$ Solution
$\displaystyle \int_{1}^{4}{{2{z^5}\sqrt {2 + 9{z^2}} \,dz}}$ Solution
$\displaystyle \int{{\frac{1}{{\sqrt {9{x^2} - 36x + 37} }}\,dx}}$ Solution
$\displaystyle \int{{\frac{{{{\left( {z + 3} \right)}^5}}}{{{{\left( {40 - 6z - {z^2}} \right)}^{\frac{3}{2}}}}}\,dz}}$ Solution
$\displaystyle \int{{\cos \left( x \right)\sqrt {9 + 25 \sin^2\left( x \right)} \,dx}}$ Solution