Section 5.3 : Substitution Rule for Indefinite Integrals
6. Evaluate \( \displaystyle \int{{\sec \left( {1 - z} \right)\tan \left( {1 - z} \right)\,dz}}\).
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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.
Start Solution
In this case it looks like we should use the following as our substitution.
\[u = 1 - z\]
Hint : Recall that after the substitution all the original variables in the integral should be replaced with \(u\)’s.
Show Step 2
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.
\[du = - dz\]To help with the substitution let’s do a little rewriting of this to get,
\[dz = - du\]Show Step 3
Doing the substitution and evaluating the integral gives,
\[\int{{\sec \left( {1 - z} \right)\tan \left( {1 - z} \right)\,dz}} = - \int{{\sec \left( u \right)\tan \left( u \right)\,du}} = - \sec \left( u \right) + c\]
Hint : Don’t forget that the original variable in the integrand was not \(u\)!
Show Step 4
Finally, don’t forget to go back to the original variable!
\[\int{{\sec \left( {1 - z} \right)\tan \left( {1 - z} \right)\,dz}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \sec \left( {1 - z} \right) + c}}\]