Question

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

Answer 1

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`