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Section 5.3 : Substitution Rule for Indefinite Integrals
11. Evaluate \( \displaystyle \int{{4\left( {\frac{1}{z} - {{\bf{e}}^{ - z}}} \right)}}\cos \left( {{{\bf{e}}^{ - z}} + \ln z} \right)\,dz\).
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Hint : Recall that if there is a term in the integrand (or a portion of a term) with an “obvious” inside function then there is at least a chance that the “inside” function is the substitution that we need.
In this case it looks like we should use the following as our substitution.
\[u = {{\bf{e}}^{ - z}} + \ln z\]
Hint : Recall that after the substitution all the original variables in the integral should be replaced with \(u\)’s.
Because we need to make sure that all the \(z\)’s are replaced with \(u\)’s we need to compute the differential so we can eliminate the \(dz\) as well as the remaining \(z\)’s in the integrand.
\[du = \left( { - {{\bf{e}}^{ - z}} + \frac{1}{z}} \right)dz\]Recall that, in most cases, we will also need to do a little manipulation of the differential prior to doing the substitution. In this case we don’t need to do that.
Show Step 3Doing the substitution and evaluating the integral gives,
\[\int{{4\left( {\frac{1}{z} - {{\bf{e}}^{ - z}}} \right)\cos \left( {{{\bf{e}}^{ - z}} + \ln z} \right)\,dz}} = \int{{4\cos \left( u \right)\,du}} = 4\sin \left( u \right) + c\]
Hint : Don’t forget that the original variable in the integrand was not \(u\)!
Finally, don’t forget to go back to the original variable!
\[\int{{4\left( {\frac{1}{z} - {{\bf{e}}^{ - z}}} \right)\cos \left( {{{\bf{e}}^{ - z}} + \ln z} \right)\,dz}} = \require{bbox} \bbox[2pt,border:1px solid black]{{4\sin \left( {{{\bf{e}}^{ - z}} + \ln z} \right) + c}}\]