Solving ∫dx/a^2sin^2x+b^2cos^2x | Homework Equations & Attempt at Solution

[Dumbledore211]

Homework Statement

∫dx/a^2sin^2x+b^2cos^2x

Homework Equations

The Attempt at a Solution

Okay, I attempted to solve this integral analytically but I couldn't. I tried to solve this by using trig identities like (1+cos2x)/2 and (1-cos2x)/2 but it got even more complicated. Can any of you drop a hint as to how it should be done

 

[vela]

You can rewrite the denominator as $\cos^2 x\ (a^2 \tan^2 x + b^2)$. This suggests the substitution u = tan x. Try that.

Or try rational trig substitutions. http://www.sosmath.com/calculus/integration/raextrig/raextrig.html

 

[Dumbledore211]

The whole denominator thing seems very complex for me to solve. How can I convert the denominator into a solvable numerator considering that the cos^2x is linked with b^2. Even if I take the U substitution route by considering u=tanx how do i deal with b^2cos^2x. How about I use the trig identity tan(x/2)/1+tan^2x/2=sinx and cosx= 1-tan^x/2/1+tan^x/2 then consider u=tanx/2 then we have it's derivative 1(1+tan^2x/2)/2=sec^2x/2/2

 

[LCKurtz]

You can rewrite the denominator as $\cos^2 x\ (a^2 \tan^2 x + b^2)$. This suggests the substitution u = tan x. Try that.

Or try rational trig substitutions. http://www.sosmath.com/calculus/integration/raextrig/raextrig.html

Or $b^2\cos^2(x)(\frac {a^2}{b^2}\tan^2x + 1)$ and let $u = \frac a b \tan x$. Guaranteed.