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February 18, 2026
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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In this case it looks like we should use the following as our substitution.
\[u = 1 - z\]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 3Doing the substitution and evaluating the integral gives,
\[\int{{\sec \left( {1 - z} \right)\tan \left( {1 - z} \right)\,dz}} = - \int{{\sec \left( u \right)\tan \left( u \right)\,du}} = - \sec \left( u \right) + c\]Finally, don’t forget to go back to the original variable!
\[\int{{\sec \left( {1 - z} \right)\tan \left( {1 - z} \right)\,dz}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \sec \left( {1 - z} \right) + c}}\]