Math  /  Calculus

QuestionEvaluate the definite integral. 181x16x23dx\int_{1}^{8} \frac{1}{x \sqrt{16 x^{2}-3}} d x

Studdy Solution
Evaluate the definite integral using the new limits:
[lntan(θ)+tan2(θ)2]sec1(4)sec1(32) \left[ \ln \left| \tan(\theta) + \sqrt{\tan^2(\theta) - 2} \right| \right]_{\sec^{-1}(4)}^{\sec^{-1}(32)}
Calculate the values at the limits and subtract:
lntan(sec1(32))+tan2(sec1(32))2lntan(sec1(4))+tan2(sec1(4))2 \ln \left| \tan(\sec^{-1}(32)) + \sqrt{\tan^2(\sec^{-1}(32)) - 2} \right| - \ln \left| \tan(\sec^{-1}(4)) + \sqrt{\tan^2(\sec^{-1}(4)) - 2} \right|
The value of the definite integral is:
lntan(sec1(32))+tan2(sec1(32))2tan(sec1(4))+tan2(sec1(4))2 \boxed{\ln \left| \frac{\tan(\sec^{-1}(32)) + \sqrt{\tan^2(\sec^{-1}(32)) - 2}}{\tan(\sec^{-1}(4)) + \sqrt{\tan^2(\sec^{-1}(4)) - 2}} \right|}

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