Calculus II - Integrals Involving Trig Functions

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Section 7.2 : Integrals Involving Trig Functions

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13. Evaluate \( \displaystyle \int{{\frac{{2 + 7{{\sin }^3}\left( z \right)}}{{{{\cos }^2}\left( z \right)}}\,dz}}\).

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Because of the sum in the numerator it makes some sense (hopefully) to maybe split the integrand (and then the integral) up into two as follows.

\[ \int{{\frac{{2 + 7{{\sin }^3}\left( z \right)}}{{{{\cos }^2}\left( z \right)}}\,dz}} = \int{{\frac{2}{{{{\cos }^2}\left( z \right)}} + \frac{{7{{\sin }^3}\left( z \right)}}{{{{\cos }^2}\left( z \right)}}\,dz}} = \int{{\frac{2}{{{{\cos }^2}\left( z \right)}}\,dz}} + \int{{\frac{{7{{\sin }^3}\left( z \right)}}{{{{\cos }^2}\left( z \right)}}\,dz}}\] Show Step 2

Now, the first integral looks difficult at first glance, but we can easily rewrite this in terms of secants at which point it becomes a really easy integral.

For the second integral again, think about how we would do that if it was a product instead of a quotient. In that case we would simply strip out a sine.

\[ \int{{\frac{{2 + 7{{\sin }^3}\left( z \right)}}{{{{\cos }^2}\left( z \right)}}\,dz}} = \int{{2{{\sec }^2}\left( z \right)\,dz}} + 7\int{{\frac{{{{\sin }^2}\left( z \right)}}{{{{\cos }^2}\left( z \right)}}\sin \left( z \right)\,dz}}\] Show Step 3

As noted above the first integral is now very easy (which we’ll do in the next step) and for the second integral we can use the trig identity \({\sin ^2}\theta + {\cos ^2}\theta = 1\) to convert the remaining sines in the second integral to cosines.

\[ \int{{\frac{{2 + 7{{\sin }^3}\left( z \right)}}{{{{\cos }^2}\left( z \right)}}\,dz}} = \int{{2{{\sec }^2}\left( z \right)\,dz}} + 7\int{{\frac{{1 - {{\cos }^2}\left( z \right)}}{{{{\cos }^2}\left( z \right)}}\sin \left( z \right)\,dz}}\] Show Step 4

Now we can use the substitution \(u = \cos \left( z \right)\) to evaluate the second integral. The first integral doesn’t need any extra work.

\[\begin{align*}\int{{\frac{{2 + 7{{\sin }^3}\left( z \right)}}{{{{\cos }^2}\left( z \right)}}\,dz}} & = 2\tan \left( z \right) - 7\int{{\frac{{1 - {u^2}}}{{{u^2}}}\,du}}\\ & = 2\tan \left( z \right) - 7\int{{{u^{ - 2}} - 1\,du}} = 2\tan \left( z \right) - 7\left( { - {u^{ - 1}} - u} \right) + c\end{align*}\]

Note that we’ll not be doing the actual substitution work here. At this point it is assumed that you recall substitution well enough to fill in the details if you need to. If you are rusty on substitutions you should probably go back to the Calculus I practice problems and practice on the substitutions.

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Don’t forget to substitute back in for \(u\)!

\[ \int{{\frac{{2 + 7{{\sin }^3}\left( z \right)}}{{{{\cos }^2}\left( z \right)}}\,dz}} = \require{bbox} \bbox[2pt,border:1px solid black]{{2\tan \left( z \right) + 7\frac{1}{{\cos \left( z \right)}} + 7\cos \left( z \right) + c = 2\tan \left( z \right) + 7\sec \left( z \right) + 7\cos \left( z \right) + c}}\]