Integral of cos(t)/(1+9sin^2(t)) Explained

  • Thread starter Thread starter americanforest
  • Start date Start date
  • Tags Tags
    Integral
americanforest
Messages
220
Reaction score
0
Hey guys,

By what rules is the integral of

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

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 u = 3\sin t

and use the fact that \int \frac{dx}{x^{2}+a^{2}} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C
 
Last edited:
gotcha, thanks
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top