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Antiderivative of trig function

  • Thread starter nuuskur
  • Start date
  • #1
523
342

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.
 

Answers and Replies

  • #2
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Multiply by [itex] \frac{sec^2x}{sec^2x} [/itex] and then substitute [itex] y=\sqrt 2 \tan x [/itex]!
 
  • #3
523
342
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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