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Evaluate the integral.
$ \displaystyle \int \tan^3 x \sec^6 x dx $
$$\frac{1}{8} \tan ^{8} x+\frac{1}{3} \tan ^{6} x+\frac{1}{4} \tan ^{4} x+C$$
Calculus 2 / BC
Chapter 7
Techniques of Integration
Section 2
Trigonometric Integrals
Integration Techniques
Campbell University
Harvey Mudd College
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this problem is from Chapter seven section to problem number twenty nine in the book Calculus Early Transcendental lt's a condition by James Door. We have an indefinite integral of tangent Cube time seeking to the sixth power Since we haven't even power on the sea can't let's proceed by pulling out, seek and squared We have tangent cute times he can to the fourth which weaken Raya's c can squared square So this is our Sikh into the fourth right here and we still have our remaining seek and squared at the end The reason for pulling out the sea can't square so that if we do it you sub you equal standing Then we see that do you will be seeking square eggs d x So if we want to use the u sub, we'Ll also have to deal with the Sikh hands squared and the parentheses And the way to deal with this and the parentheses is to apply the path Agron identity seek and squared is tan squared plus one. So let's use this identity. Next we have tank you and then in the parentheses applying this But there is an identity in the right. We have tan squared X plus one in the Prentice's That's all being squared. Time sequence Word. Now, at this point, we see that we can apply the U substitution u equals tangent of X. So if we do this so tan Cube becomes you Cube And in the parentheses we have you square plus one that's all squared in the second square x t x That's just do you. So let's go ahead and simplify this Interbrand whether you cued and then in the parentheses, we have you forth plus two, you square plus one and let's go ahead and distribute issue to the third power through the apprentices so that we get you to the seventh to you to the fifth. Plus you, kun, do you and we could evaluate these three intervals using the power rule. We have you to the eight over eight to you to the six over six plus you know, the fourth over four and plus our constant C. And finally we can write our final answer in terms of X by going back started you sub So let's replace you with Tangent of X and we have you to the eighth becomes tangent to the power of X. And we could simplify this fraction too. Over sixes, one over three. So we have tangents to the sixth Power Vicks Laboratory, and then we have you to the fourth over four. So that becomes attention to the fourth all over for you and plus their constancy, and that's our answer.
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