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Evaluate the integral.
$ \displaystyle \int \tan^5 x \sec^3 x dx $
$\frac{\sec ^{7}(x)}{7}-\frac{2 \sec ^{5}(x)}{5}+\frac{\sec ^{3}(x)}{3}+C$
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 eight in the book Calculus Early Transcendental lt's a Condition by James Store Here we have an indefinite integral of tangent to the fifth power of ext Time seeking cute of X Since we have an odd power on the sea can't let's proceed by pulling out a factor of tangent times he can't We have tangent to the fourth power times seek and square And then we have our remaining factors of tangent of X time seeking a Vicks No. So the reason for doing this is that we could eventually use the U substitution u equals seek innovex If we do that with the DOAs seeking of X times ten eggs d x and that's exactly the term that we've underlined. So we see that this is the use of that will want This will be you squared But we'll have to deal with this Tan's into the fourth power term in orderto express this in terms of you So let's go to the side and deal with this the scans and chance into the fourth power so we can write tangent to the fourth is tangent squared, squared, using the laws of experiments. And then we can use this property appear the protagonist identity to rewrite that as seeking square X minus one square. Good. So after we apply really substitution tangents of the fourth becomes you square minus one square We have seek into the square So that's you square and this remaining term tangent Time seeking DX That's our do you. So let's go ahead and evaluate this square first. If you know the four power minus two, you square plus one all times square So let's go ahead and now distribute this new square through the apprentices and we could use the power rule three times to evaluate each of these in A girls you to the seven over seven minus to you to the fifth over five Plus you cute number three and our constant denigration. See? And finally, we can read our answer in terms of X by going back to our U substitution. So you two the seventh becomes see cancer. The seven power vex over seven minus to seek answers with power over five plus seeking to the third power over three and plus see. And there's our answer
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$ \displaystyle \int \tan^3 x \sec x dx $
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