Evaluate the following :
`int_(0)^(pi//2)(1)/(a^(2)sin^(2)x+b^(2)cos^(2)x)dx`.

Explanation

Let `I=int_(0)^(pi//2)(1)/(a^(2)sin^(2)x+b^(2)cos^(2)x)dx`
Dividing both numerator and denominator by `cos^(2)x,` we get,
`I=int_(0)^(pi//2)(sec^(2)xdx)/(a^(2)tan^(2)x+b^(2))`
Put `tanx=t.` Then `sec^(2)xdx=dt`
When `x=0, t=tan0=0," When "x=(pi)/(2), t=tan.(pi)/(2)=oo`
`therefore" "I=int_(0)^(oo)(1)/(a^(2)t^(2)+b^(2))dt=(1)/(a^(2))int_(0)^(oo)(1)/(t^(2)+(b//a)^(2))dt`
`=(1)/(a^(2))xx(1)/((b//a))[tan^(-1)((t)/(b//a))]_(0)^(oo)`
`=(1)/(ab)[tan^(-1)((at)/(b))]_(0)^(oo)`
`=(1)/(ab)[tan^(-1)oo-tan^(-1)0]" "=(1)/(ab)[(pi)/(2)-0]=(pi)/(2ab).`