Section 5.8 : Substitution Rule for Definite Integrals
Evaluate each of the following integrals, if possible. If it is not possible clearly explain why it is not possible to evaluate the integral.
- \( \displaystyle \int_{{ - 2}}^{3}{{\frac{4}{{{{\left( {5 + 2x} \right)}^3}}}\,dx}}\)
- \( \displaystyle \int_{1}^{0}{{10\left( {1 - 2{w^2}} \right)\sqrt[4]{{7 - 3w + 2{w^3}}}\,dw}}\)
- \( \displaystyle \int_{{ - 1}}^{4}{{\left( {t - 2} \right){{\bf{e}}^{{t^{\,2}} - 4t}}\,dt}}\)
- \( \displaystyle \int_{1}^{6}{{7\cos \left( {\frac{{\pi z}}{2}} \right){{\left( {4 + \sin \left( {\frac{{\pi z}}{2}} \right)} \right)}^5}\,dz}}\)
- \( \displaystyle \int_{0}^{1}{{\frac{{{w^3}}}{{6{w^4} + 3}}\,dw}}\)
- \( \displaystyle \int_{{ - 1}}^{1}{{{x^2}\cos \left( {{x^3} + 2} \right) - {x^2}{{\bf{e}}^{{x^3} + 2}}\,dx}}\)
- \( \displaystyle \int_{0}^{{\frac{\pi }{3}}}{{\frac{{4\sin \left( {3t} \right)}}{{2 + \cos \left( {3t} \right)}} + \frac{{7\sin \left( {3t} \right)}}{{{{\left( {2 + \cos \left( {3t} \right)} \right)}^2}}}\,dt}}\)
- \( \displaystyle \int_{0}^{\pi }{{{{\sec }^2}\left( y \right)\sqrt {2 + \tan \left( y \right)} \,dy}}\)
- \( \displaystyle \int_{1}^{9}{{\sqrt {{x^5}} + \frac{{\sin \left( {\sqrt x } \right)}}{{\sqrt x }}\,dx}}\)
- \( \displaystyle \int_{0}^{1}{{{{\sec }^2}\left( w \right) - \frac{2}{{4{w^2} + 1}}dw}}\)
- \( \displaystyle \int_{3}^{0}{{{{\bf{e}}^{ - 4t}}\sqrt {2 + {{\bf{e}}^{ - 4t}}} + 8{{\bf{e}}^t}\,dt}}\)
- \( \displaystyle \int_{3}^{7}{{\frac{{9{{\bf{e}}^x}}}{{{{\bf{e}}^x} + 4}} + \frac{{{{\left[ {\ln \left( {2x} \right)} \right]}^2}}}{x}\,dx}}\)
- \( \displaystyle \int_{0}^{\pi }{{\sin \left( {\frac{v}{2}} \right)\left[ {6 + 3{{\cos }^2}\left( {\frac{v}{2}} \right) - 4{{\cos }^4}\left( {\frac{v}{2}} \right)} \right]\,dv}}\)
- \( \displaystyle \int_{1}^{2}{{{{\bf{e}}^{ - t}} + 3t{{\bf{e}}^{5 - {t^2}}}\,dt}}\)
- \( \displaystyle \int_{0}^{6}{{\frac{{8{t^3}}}{{2{t^4} + 1}} - \frac{{7t}}{{{t^2} - 9}}\,dt}}\)
- \( \displaystyle \int_{2}^{6}{{\sqrt {1 + 2y} + \left( {4 - y} \right){{\left( {{y^2} - 8y + 5} \right)}^4}\,dy}}\)
- \( \displaystyle \int_{0}^{1}{{{{\bf{e}}^{2z}}\sin \left( {{{\bf{e}}^{2z}} - 1} \right) + \sin \left( z \right){{\bf{e}}^{2 - \cos \left( z \right)}}\,dz}}\)