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Any hints of how to approach solving this integral?

$$\int_{0}^{2\pi} \frac{dx}{a^2\sin^2x+b^2\cos^2x}, \quad a,b>0$$

I tried substituting $z=e^{ix}$ but I get a rather complex form I can't do anything with.

tyr
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  • have you tried using $sin^2 x =1 - cos^2 x$ in the denominator, then collecting like terms? – MKF Jan 20 '18 at 09:40
  • Then, since there is the appearance of a trigonometric function, use the universal substitution $t=\tan(x)$ (or $t=\tan(x/2)$ if you prefer). – MKF Jan 20 '18 at 09:46
  • Also related: https://math.stackexchange.com/questions/518173/calculate-the-integral-int-02-pi-frac1a2-cos2tb2-sin2tdt. – Martin R Jan 20 '18 at 09:54
  • Hint: Enforcing the change of variables $x= t-π$ lead to

    $$\int_{0}^{2\pi} \frac{dx}{a^2\sin^2x+b^2\cos^2x}=2\int_{0}^{\pi} \frac{dx}{a^2\sin^2x+b^2\cos^2x}=4\int_{0}^{\pi/2} \frac{dx}{a^2+(b^2-a^2)\cos^2x},$$

    then use $u=tan(x)$

     – Guy Fsone Jan 20 '18 at 10:03

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Try converting the limit to $ 0 $ to $\pi/2$ and then split it into two integral of limits $ 0 $ to $\pi/4$ and $ \pi/4$ to $\pi/2$ . Substitute u=tan x in first integral and v=cot x in second.

Reader Manifold
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