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JH
Numerade Educator

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Problem 22 Easy Difficulty

Evaluate the integral.

$ \displaystyle \int \tan^2 \theta \sec^4 \theta d \theta $

Answer

$\frac{\tan ^{3} \theta}{3}+\frac{\tan ^{5} \theta}{5}+C$

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Video Transcript

this problem is from chapter seven, Section two. Problem number twenty two in the book Calculus Early Transcendental. Lt's a condition by James Store Here we have an indefinite integral of Tan Xin Square of St Time. See cancer, the fourth power of data. So the first thing we can do here is simply rewrite this. Seeking to the fourth power a C can't square of time sequence Where? So doing this. We have ten square data times sequence where data seeking squared data d theta and we can apply this identity over here in the blue to replace. He can't squared with tan squared plus one. Let's do this for one of the Sikh and squares. Let's not do it for both so that we can use the U substitution. Now we see we can apply u substitution here, Let's take you two be tan data so that do you is seeking square data data applying this u substitution we see here we have do you useful here and in the parentheses to use where plus one so in general should become use where times you squared plus one deal. So before we evaluate integral, that's just multiply this out, we get you to the fourth Power. Plus you square, do you? And we can apply the power rule twice to evaluate these two. We obtained you to the fifth power over five, plus you to the third power over three and then our constancy of integration. And for the last step, we come back to our U substitution toe back substitute from you back into a tan. Data Silly obtained ten to the fifth hour, two five plus ten to the third power of data. Plus he and that's our answer.