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Evaluate the integral.

$ \displaystyle \int \tan^3 x \sec x dx $

$$\frac{1}{3} \sec ^{3} x-\sec x+C$$

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 2

Trigonometric Integrals

Integration Techniques

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this problem is from Chapter seven section to problem number twenty seven in the book Calculus Early Transcendental lt's a Tradition by James Store We haven't indefinite inaugural of Tangent Cube time, so you can't since the power on seek an Izod is just one here. Let's pull out a factor of tangent times he can outside of this inner grant. So I have tangent Squared X And then here's the other power of tangent Times are seeking and the reason for doing so is that we'll eventually be able to deal with this term tangent time c can when we do it u substitution and that you saw it should be u equals seek an ibex But before we do that, we'Ll have to deal with this tan squared over here on the left So let's rewrite this tan squared Soto, rewrite this tan square We can use our identity Our protagonist identity on the right Over here circled c can square is tan squared plus one So that means town square is C can't squared minus one So that becomes That's our chance Weird and we still have this remaining term left over. Now we see that we have the right choice for our U substitution. Let's take you to be seeking so that do you is seek Innovex Time's tangent of X dx. So if we come backto are integral we see that here we have in terms of our use of we have a use squared minus one and the remaining term over here is our do you So integral becomes you squared minus one. Do you? And we can evaluate each of these inaugurals by using the power rule. So you swear becomes you, keep over three and one just becomes you and lets out our constancy of integration. And at this point, we can We're done with the integration. But we should put our final answer in terms of X. So we come back to the U substitution to rewrite this. And so you cute becomes seeking kun the Vex mine issue, which is just seeking a vex plus C. And there's our answer

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