Calculus I - Substitution Rule for Indefinite Integrals (Practice Problems)

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Section 5.3 : Substitution Rule for Indefinite Integrals

For problems 1 – 16 evaluate the given integral.

$\displaystyle \int {{\left( {8x - 12} \right){{\left( {4{x^2} - 12x} \right)}^4}\,dx}}$ Solution
$\displaystyle \int {{3{t^{ - 4}}{{\left( {2 + 4{t^{ - 3}}} \right)}^{ - 7}}\,dt}}$ Solution
$\displaystyle \int {{\left( {3 - 4w} \right){{\left( {4{w^2} - 6w + 7} \right)}^{10}}\,dw}}$ Solution
$\displaystyle \int {{5\left( {z - 4} \right)\,\,\,\sqrt[3]{{{z^2} - 8z}}\,dz}}$ Solution
$\displaystyle \int {{90{x^2}\sin \left( {2 + 6{x^3}} \right)\,dx}}$ Solution
$\displaystyle \int {{\sec \left( {1 - z} \right)\tan \left( {1 - z} \right)\,dz}}$ Solution
$\displaystyle \int {{\left( {15{t^{ - 2}} - 5t} \right)\cos \left( {6{t^{ - 1}} + {t^2}} \right)\,dt}}$ Solution
$\displaystyle \int {{\left( {7y - 2{y^3}} \right){{\bf{e}}^{{y^{\,4}} - 7{y^{\,2}}}}\,dy}}$ Solution
$\displaystyle \int {{\frac{{4w + 3}}{{4{w^2} + 6w - 1}}\,dw}}$ Solution
$\displaystyle \int {{\left( {\cos \left( {3t} \right) - {t^2}} \right){{\left( {\sin \left( {3t} \right) - {t^3}} \right)}^5}\,dt}}$ Solution
$\displaystyle \int {{4\left( {\frac{1}{z} - {{\bf{e}}^{ - z}}} \right)}}\cos \left( {{{\bf{e}}^{ - z}} + \ln z} \right)\,dz$ Solution
$\displaystyle \int {{{{\sec }^2}\left( v \right){{\bf{e}}^{1 + \tan \left( v \right)}}\,dv}}$ Solution
$\displaystyle \int {{10\sin \left( {2x} \right)\cos \left( {2x} \right)\sqrt {{{\cos }^2}\left( {2x} \right) + 5} \,dx}}$ Solution
$\displaystyle \int {{\frac{{\csc \left( x \right)\cot \left( x \right)}}{{2 - \csc \left( x \right)}}\,dx}}$ Solution
$\displaystyle \int {{\frac{6}{{7 + {y^2}}}\,dy}}$ Solution
$\displaystyle \int {{\frac{1}{{\sqrt {4 - 9{w^2}} }}\,dw}}$ Solution
Evaluate each of the following integrals.
1. $\displaystyle \int {{\frac{{3x}}{{1 + 9{x^2}}}\,dx}}$
2. $\displaystyle \int {{\frac{{3x}}{{{{\left( {1 + 9{x^2}} \right)}^4}}}\,dx}}$
3. $\displaystyle \int {{\frac{3}{{1 + 9{x^2}}}\,dx}}$
Solution