Integral derivation: int[0,pi/2][(a^2cos^2x+b^2sin^2x)^-2]dx

[micke]

Hi,
Could someone help with this problem? I don't even know where to start.

Calculate for a>0 and b>0 the integral

\(\displaystyle \int\limits_0^{\pi /2} {(a^2 \cos ^2 x + b^2 \sin ^2 x)^{ - 2} dx}\)

with help from the partial derivatives with respect to a and b of

\(\displaystyle \int\limits_0^{\pi /2} {(a^2 \cos ^2 x + b^2 \sin ^2 x)^{ - 1} dx}\)

I've calculated these derivatives

\(\displaystyle \int\limits_0^{\pi /2} { - 2a*\cos ^2 x(a^2 \cos ^2 x + b^2 \sin ^2 x)^{ - 2} dx}\)

and

\(\displaystyle \int\limits_0^{\pi /2} { - 2b*\sin ^2 x(a^2 \cos ^2 x + b^2 \sin ^2 x)^{ - 2} dx}\)

but I don't know how I'm supposed to make use of them to solve the problem.

 

[royhaas]

Perhaps you are supposed to evaluate the integral
\(\displaystyle \int\limits_0^{\pi /2} {(a^2 \cos ^2 x + b^2 \sin ^2 x)^{ - 1} dx}\)
first, for example, using the substitution \(\displaystyle u=\tan(x/2)\). This should transform it into the integral of a rational function over the unit interval. Then take the partial derivatives. Or look up the value of that integral in a table of definite integrals.

 

[Dr. Flim-Flam]

Integral[a^2(cos(x))^2+b^2(sin(x))^2]^-2,x,0,Pi/2 = Pi(a^2+b^2)/(4a^3b^3), a and b >0. (TI-89)