Why does ∫cos(t)/(1+9sin²(t))dt = (1/3)tan⁻¹(3sin(t))?

  • Thread starter Thread starter americanforest
  • Start date Start date
  • Tags Tags
    Integral
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 2K views
americanforest
Messages
220
Reaction score
0
Hey guys,

By what rules is the integral of

[tex]\frac{cos(t)}{1+9sin^2(t)}=\frac{1}{3}tan^{-1}(sin(3t))[/tex]

I know this is right, but I have no idea why and can't find any trig identities to help. Thanks.
 
Physics news on Phys.org
Let [tex]u = 3\sin t[/tex]

and use the fact that [tex]\int \frac{dx}{x^{2}+a^{2}} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C[/tex]
 
Last edited:
gotcha, thanks