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

Putting `x=asin theta` we get `dx=a cos theta d theta`.
When `x=0=asin theta, theta=0`
When `x=a=asintheta, sin theta =1`. Therefore `theta=pi//2`
The given integral
`I=int_(0)^(pi//2)(cos theta d theta)/(sin theta +cos theta)`……….....1
`:.I=int_(0)^(pi//2)(cos(1/2pi-theta)d theta)/(sin(1/2pi - theta)+cos (1/2pi - theta))`
`=int_(0)^(pi//2) (sin theta d theta)/(cos theta +sin theta)`
Adding 1 and 2, we get
`2I=int_(0)^(pi//2)(sin theta+cos theta)/(sin theta+cos theta) d theta=int_(0)^(pi//2) d theta`
`=[theta]_(0)^(pi//2)=pi//2`
`:. I=pi//4`