Another couple of integration porblems

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SUMMARY

The discussion focuses on integrating two complex expressions involving hyperbolic and trigonometric functions. The first integral, \(\int \text{sech}\left(\frac{1}{x}\right)\tanh\left(\frac{1}{x}\right)\frac{1}{x^2}dx\), can be approached by recognizing the property \(\int \text{tanh}(x)\text{sech}(x)dx = -\text{sech}(x) + C\). The second integral, \(\int \cos^{-3}(2\theta)\sin(2\theta)d\theta\), can be simplified using the substitution \(u = \cos(2\theta)\). Both substitutions clarify the integration process significantly.

PREREQUISITES
  • Understanding of hyperbolic functions, specifically \(\text{sech}\) and \(\tanh\).
  • Familiarity with trigonometric functions and their integrals, particularly \(\sin\) and \(\cos\).
  • Knowledge of integration techniques, including substitution methods.
  • Basic algebraic manipulation skills, such as rewriting expressions and applying properties of exponents.
NEXT STEPS
  • Study the properties and integrals of hyperbolic functions, focusing on \(\text{sech}\) and \(\tanh\).
  • Learn advanced integration techniques, including integration by substitution and integration of inverse trigonometric functions.
  • Practice integrating complex trigonometric expressions, particularly those involving powers and double angles.
  • Explore the use of substitution in integration to simplify complex integrals effectively.
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Students and educators in calculus, particularly those focusing on integration techniques involving hyperbolic and trigonometric functions.

chubb rock
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Back for a couple more I'm having a little trouble with. The first is

sech(1/x)tanh(1/x) all over x^2

Here I'm not sure how to deal with integrating hyperbolics at all as we haven't gotten to it in class yet. I did some searching on the internet and found that the integral tanh(x)sech(x) = -sech(x) + C. With all of it being over x^2 it becomes a little more complicated.

Do I rewrite x^2 as x^-2 and multiply it through with the 1/x's? Would the property tanh(x)sech(x) = -sech(x) + C still work?

I was also wondering about rewriting the whole equation as (2/e^1/x + e^-1/x)(e^1/x - e^-1/x /e^1/x - e^-1/x)(1/x^2). That seems overly complicated though and I'm not sure what form to go to from that point.

The other problem I'm having trouble with is integrating

cos^-3(2θ)sin(2θ)dθ

I know of ways of integrating sin and cos that have squares but is there a property I can use that's cubed and a double angel?
 
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What do you mean by integrating all over x^2? I suppose you mean \int \text{sech}\left(\frac{1}{x}\right)\tanh\left(\frac{1}{x}\right)\frac{1}{x^2}dx. If so use the substitution u=sech \left(\frac{1}{x}\right). For your second problem use the substitution u=\cos 2\theta
 
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Yeah, that is what I meant and both make so much more sense to me now. Thanks!
 

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