`int_(0)^(pi/2) (cos^(2)x dx)/(cos^(2)x+4 sin^(2)x)`

Explanation

`" Let I "= int_(0)^(pi//2) (cos^(2)x)/(cos^(2)x+4sin^(2)x)dx.....(1)`
`=int_(0)^(pi//2) (cos^(2)x)/(4-3cos^(2)x)dx`
`( :' sin^(2) x=1-cos^(2)x)`
`=-(1)/(3)int_(0)^(pi//2) (4-3 cos^(2) x-4)/(4-3 cos^(2)x) dx`
`=-(1)/(3) int_(0)^(pi//2) (1-(4)/(4-3 cos^(2)x))dx`
`=-(1)/(3)[x]_(0)^(pi//2) +(4)/(3) int_(0)^(pi//2) (sec^(2)x)/(4sec^(2) x-3) dx`
(Divide numerator nad denominator by `cos^(2)` x in second integral)
`=(pi)/(6)+(4)/(3) int_(0)^(pi//2) (sec^(2) x)/(4(1-tan^(2) x)-3)dx`
`underset(" and " x=((pi)/(2)) rArr t=oo)underset(" and " x=0 rArr t=0)("Let " " " tan x=t)`
`:. I= -(pi)/(6)+(4)/(3) int_(uu)^(oo) (dt)/(4(1+t^(2))-3)`
`=-(pi)/(6)+(4)/(3) int_(0)^(oo) (dt)/(4t^(2)+1)`
`=-(pi)/(6)+(4)/(3)xx(1)/(4) int_(0)^(oo)(dt)/(t^(2)+((1)/(2))^(2))`
`=-(pi)/(6) +(1)/(3) .2 [tan^(-1) .((t)/(1))/((1)/(2))]_(0)^(oo)`
`=-(pi)/(6)+(2)/(3) [tan^(-1) oo-tan^(-1) 0)`
`=-(pi)/(6)+(2)/(3) ((pi)/(2)-0)=(pi)/(6)+(pi)/(3)=(pi)/(6)`