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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 $
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Calculus 2 / BC
Chapter 7
Techniques of Integration
Section 6
Integration Using Tables and Computer Algebra Systems
Integration Techniques
Harvey Mudd College
University of Michigan - Ann Arbor
University of Nottingham
Lectures
01:53
In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.
27:53
In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.
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Use the Table of Integrals…
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$5-32$ Use the Table of In…
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Use the Table of Integral…
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Use the Substitution Formu…
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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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