Evaluate: `int_0^(pi//2)1/((a^2cos^2x+b^2sin^2x)^2)dx`

Answer

N/a

Explanation

Let `I = int_(0)^(pi//2) (dx)/((a^(2)cos^(2)x+b^(2)sin^(2)x)^(2))`
Divide numerator and denominator by `cos^(4)x`, we get
`I = int_(0)^(pi//2) (sec^(4)xdx)/((a^(2)+b^(2) tan^(2)x)^(2))`
`= int_(0)^(pi/2)((1+tan^(2)x)sec^(2)xdx)/((a^(2)+b^(2)tan^(2)x)^(2))`
Put `tanx = t`
`rArr sec^(2)dx = dt`
As `x rarr 0`, then `t rarr 0`
and `x rarr pi/2`, then `t rarr oo , I = int_(0)^(oo) ((1+t^(2)))/((a^(2)+b^(2)t^(2))^(2))`
Now, `(1+t^(2))/((a^(2)+b^(2)t^(2))^(2)) , ["let" t^(2) = u]`
`(1+u)/((a^(2)+b^(2)u^(2))) = (A)/((a^(2)+b^(2)u)) + (B)/((a^(2)+b^(2)u)^(2)) `
`rArr 1+u = A(a^(2)+b^(2)u)+B`
On comparing the coefficient of x and constant term on both sides, we get
`a^(2)A + B = 1`
and ` b^(2)A = 1"....."(i)`
`A = 1/(b^(2))"......."(i)`
Now `(a^(2))/(b^(2)) + B = 1`
`B = 1 - (a^(2))/(b^(2)) = (b^(2)-a^(2))/(b^(2))`
`:. I = int_(0)^(oo) ((1+t^(2)))/((a^(2)+b^(2)t^(2))^(2))`
`= 1/(b^(2)) int_(0)^(oo) (dt)/(a^(2)+b^(2)t^(2)) + (b^(2)-a^(2))/(b^(2)) int_(0)^(oo) (dt)/((a^(2)+b^(2)t^(2))^(2))`
`= 1/(b^(2))int_(0)^(oo) (dt)/(b^(2)(a^(2)/b^(2)+t^(2)))+(b^(2)-a^(2))/(b^(2))int_(0)^(oo)(dt)/((a^(2)+b^(2)t^(2))^(2))`
`= 1/(ab^(3))[tan^(-1)((tb)/(a))]_(0)^(oo) + (b^(2)-a^(2))/(b^(2))(pi/(4).(1)/(a^(3)b))`
`= 1/(ab^(3)) [tan^(-1)oo - tan^(-1)0]+(pi)/(4).(b^(2)-a^(2))/((a^(3)b^(3)))`
`= (pi)/(2ab^(3))+(pi)/(4).(b^(2)-a^(2))/((a^(3)b^(3)))`
` = pi ((2a^(2)+b^(2)-a^(2))/(4a^(3)b^(3))) = pi/4((a^(2)+b^(2))/(a^(3)b^(3)))`