Antiderivative of trig function

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SUMMARY

The discussion focuses on finding the antiderivative of the function \(\int\frac{{\rm{d}}x}{1+2\sin^2 (x)}\). Participants suggest using the substitution \(y = \sqrt{2}\tan(x)\) to simplify the integral. By multiplying by \(\frac{\sec^2(x)}{\sec^2(x)}\), the integral transforms into \(\int\frac{\sec^2(x){\rm{d}}x}{3\tan^2(x)+1}\), which leads to a solvable form. This method effectively resolves the initial complication encountered with the substitution of \(y\).

PREREQUISITES
  • Understanding of trigonometric identities and functions
  • Familiarity with integration techniques, particularly substitution
  • Knowledge of derivatives and their relationship to antiderivatives
  • Experience with manipulating integrals involving secant and tangent functions
NEXT STEPS
  • Study advanced integration techniques, focusing on trigonometric substitutions
  • Learn about the properties and applications of the secant function in calculus
  • Explore the derivation and applications of arctangent in integration
  • Practice solving integrals involving combinations of sine and cosine functions
USEFUL FOR

Students studying calculus, particularly those focusing on integration techniques, as well as educators looking for examples of trigonometric integrals and substitutions.

nuuskur
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Homework Statement


[tex]\int\frac{{\rm{d}}x}{1+2\sin^2 (x)}[/tex]

Homework Equations

The Attempt at a Solution


it's some sort of a derivative of arctan, however, when I try to substitute [itex]y = \sqrt{2}\sin(x),\ {\rm{d}}y = \sqrt{2}\cos(x){\rm{d}}x[/itex] I get nowhere with it, atleast I think, since there is a further complication, namely the dy. Suggestions, please.
 
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Multiply by [itex]\frac{sec^2x}{sec^2x}[/itex] and then substitute [itex]y=\sqrt 2 \tan x[/itex]!
 
Right, of course I won't think of that :D
[tex]\int\frac{\sec^2(x){\rm{d}}x}{\frac{1}{\cos^2(x)}+2\frac{\sin^2(x)}{\cos^2(x)}} = \int\frac{\sec^2(x){\rm{d}}x}{3\tan^2(x)+1}[/tex] and that's all she wrote.
 

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