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Use the Table of Integrals on Reference Pages 6-10 to evaluate the integral. $ \displaystyle \int \frac{\sec^2 \theta \tan^2 \theta}{\sqrt{9 - \tan^2 \theta}}\ d \theta $

$-\frac{\tan \theta}{2} \sqrt{9-\tan ^{2} \theta}+\frac{9}{2} \sin ^{-1}\left(\frac{\tan \theta}{3}\right)+C$

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 6

Integration Using Tables and Computer Algebra Systems

Integration Techniques

Missouri State University

Oregon State University

Harvey Mudd College

University of Nottingham

Lectures

01:11

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03:05

00:31

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Okay, so this question wants us to find the integral of this function. So to do that, let's convert it to one of the forms that we can look up in our table easily. So a rule of thumb is that we always like to have a use squared in that denominator in the radical. So let's pick you to be equal to tangent data because then we get a nine minus you square down there. So taking d'you we've see that that's C can't squared data d theta, which is convenient because we already have that. So that's all we have to D'oh! So now we can just rewrite this as the integral of well, tangents squared is you squared, Do you? Over square root of nine minus you squared. So now from here, this looks very similar to one of our forms in the table which has this formula. So in this case, A is equal to three because we have a nine, which is three squared. So now our anti derivative is negative. You over too square root nine minus you squared, plus a squared over two, which is nine divided by two times sine inverse of you divided by a, which is plus C So now all we got to Dio is plug back in tangent data for you and we're done. So we get negative. 10th 8 over to square view nine minus 10 square data plus nine halves times the inverse sine of 10th ADA divided by three plus our integration constant. And that's our final answer.

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