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### 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!

$\int{{\sec \left( {1 - z} \right)\tan \left( {1 - z} \right)\,dz}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \sec \left( {1 - z} \right) + c}}$