Evaluate: `int_0^a(dx)/(x+sqrt((a^2-x^2)))orint_0^(pi/2)(dtheta)/(1+tantheta)`

Explanation

Let `=int_(0)^(a)(1)/(xsqrt(a^(2)-x^(x))).dx`
Let `x=asintheta`
`:.dx=acostheta.d""theta`
when `x=0impliestheta=0,"when "x=aimpliestheta=(pi)/(2)`
`:.I=int_(0)^(pi//2)(acostheta)/(asintheta+sqrt(a^(2)-a^(2)sin^(2)theta))d""theta`
`=int_(0)^(pi//2)(costheta.d""theta)/(sintheta+sqrt(1-sin^(2)theta))`
`I=int_(1)^(pi//2)(costheta.d""theta)/(sintheta+costheta)" ".......(i)`
`I=int_(0)^(pi//2)(cos((pi)/(2)-theta).d""theta)/(sin((pi)/(2)-theta)+cos((pi)/(2)-theta))`
`[:'int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx]`
`I=int_(0)^(pi//2)(sintheta)/(costheta+sintheta)d""theta" "......(ii)`
On adding (i) and (ii), we get
`2I=int_(0)^(pi//2)(sintheta+costheta)/(sintheta+costheta).d""theta`
`:.2I=int_(0)^(pi//2)d""theta=[theta]_(0)^(pi//2)`
`:.2I=(pi)/(2)-0=(pi)/(2)`
`:.I=(pi)/(2)xx(1)/(2)=(pi)/(4)`