How do you find the definite integral for: dx /(a cos^2x + b sin^2 x)^2 for the intervals [0, pi]?

1 Answer
Jun 4, 2018

Assuming a and b are positive real numbers, use the substitution sqrtbtanx=sqrtatantheta.

Explanation:

Let

I=int_0^pidx/(acos^2x+bsin^2x)^2

From symmetry of the squared functions:

I=2int_0^(pi/2)dx/(acos^2x+bsin^2x)^2

Multiply numerator and denominator by sec^4x:

I=2int_0^(pi/2)(sec^4x)/(a+btan^2x)^2dx

Assuming a and b are positive real numbers, apply the substitution sqrtbtanx=sqrtatantheta:

I=2/(ab)^(3/2)int_0^(pi/2)(b+atan^2theta)/sec^2thetad theta

Simplify:

I=2/(ab)^(3/2)int_0^(pi/2)(bcos^2theta+asin^2theta)d theta

Apply the double-angle formula for cos2theta:

I=1/(ab)^(3/2)int_0^(pi/2){(a+b)-(a-b)cos2theta}d theta

Integrate directly:

I=1/(ab)^(3/2)[(a+b)theta-1/2(a-b)sin2theta]_0^(pi/2)

Hence:

I=(a+b)/(ab)^(3/2)pi/2