Evaluate the following :
`int_(0)^(a)(dx)/(x+sqrt(a^(2)-x^(2)))`.

Explanation

Let `I=int_(0)^(a)(dx)/(x+sqrt(a^(2)-x^(2)))`
Put `x=a sin theta.` Then `dx=a cos theta d theta`
and `sqrt(a^(2)-x^(2))=sqrt(a^(2)-a^(2)sin^(2)theta)=sqrt(a^(2)(1-sin^(2) theta))=a cos theta`
When x = 0 , ` a sin theta=0 " "therefore" "theta=0`
When `x=a, sin theta = a " "therefore theta=pi//2`
`therefore" "I=int_(0)^(pi//2)(a cos theta d theta)/(a sin theta+a cos theta)`
`=int_(0)^(pi//2)(cos theta d theta)/(sin theta + cos theta)`
We use the property, `int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx`
Hence in I, we change `theta ` by `(pi//2)- theta`. Then
`I=int_(0)^(pi//2)(cos[(pi//2)-theta])/(sin[(pi//2)- theta]+ cos [(pi//2)- theta])d theta`
`=int_(0)^(pi//2)(sin theta)/(cos theta+sin theta)d theta" ...(2)"`
Adding (1) and (2), we get,
`2I=int_(0)^(pi//2)(cos theta)/(cos theta+sin theta)d theta +int_(0)^(pi//2)(sin theta)/(cos theta+sin theta) d theta`
`=int_(0)^(pi//2)(cos theta+sin theta)/(cos theta +sin theta)d theta`
`=int_(0)^(pi//2)1 d theta =[ theta]_(0)^(pi//2)=(pi//2)-0 =pi//2`
`therefore I=pi//4.`