Integrals with trig substitutions =p

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SUMMARY

The discussion focuses on calculating the volume of a solid formed by revolving the region bounded by the graph of \( y = \frac{x}{\sqrt{x^2+25}} \), the x-axis, and the line \( x = 5 \) around the y-axis. The volume is determined using the shell method, represented by the integral \( V = \int 2 \pi x f(x) \). The correct answer for the volume is \( 25\pi[\sqrt{2}-\ln{\sqrt{2}+1}] \approx 41.85 \). Additionally, the user inquires about integrating \( \sec \theta \) without using a reduction formula for \( \tan^2 x \cdot \sec x \).

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  • Understanding of integral calculus and volume of solids of revolution
  • Familiarity with the shell method for volume calculation
  • Knowledge of trigonometric identities and integration techniques
  • Experience with Mathematica for computational verification
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  • Study the shell method for calculating volumes of solids of revolution
  • Learn how to integrate \( \sec \theta \) and its applications
  • Explore trigonometric substitutions in integral calculus
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Students and educators in calculus, particularly those focusing on integral techniques and volume calculations, as well as anyone using Mathematica for mathematical computations.

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NVM I GET HOW TO DO I TNOW

Homework Statement



"The region bounded by the graphs of [tex]y = \frac{x}{\sqrt{x^2+25}}[/tex], y = 0, and x = 5 is revolved about the y-axis. Find the volume of the resulting solid."

Homework Equations



The only way i see to do this right now is to use shells, and the equation for that is

[tex]V = \int 2 \pi x f(x)[/tex] where the integral is from the lower limit to the upper limit
Mathematica can't find an answer to it. You can try it yourself if you want [Integral of 2*pi*x^2/(x^2+25)^(1/2)]Thanks in advance for your helpOh also, the correct answer is [tex]25\pi[\sqrt{2}-\ln{\sqrt{2}+1}] \approx 41.85[/tex]
EDIt: ok mathematica does get an answer, but is the a way to find it without a reduction formula for tanx^2*secx ?
 
Last edited:
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I get an integrand containing

[tex]\frac{\tan^2\theta}{\sec\theta}= \frac{\sec^2\theta - 1}{\sec\theta}=\sec\theta-\cos\theta[/tex]

Do you know how to integrate [itex]\sec\theta[/itex]?
 

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