`int_(0)^(pi//2)(x)/((sinx+cosx))dx` का मान ज्ञात कीजिए ।

Explanation

माना `I=int_(0)^(pi//2)(x)/((sinx+cosx))dx" …(1)"`
` I=int_(0)^(pi//2)(((pi)/(2)-x))/(sin[(pi//2)-x]+cos[(pi//2)-x])dx`
`I=int_(0)^(pi//2)(((pi)/(2)-x))/((cosx+sinx))dx`
`=int_(0)^(pi//2)(((pi)/(2)-x))/((sinx+cosx))dx" ...(2)"`
अब, समीकरण (1 ) व (2 ), को जोड़ने पर
`@I=(pi)/(2)int_(0)^(pi//2)(dx)/((sinx+cosx))`
`I=(pi)/(4)int_(0)^(pi//2)(dx)/((sinx+cosx))`
`=(pi)/(4)int_(0)^(pi//2)(dx)/([(2tan(x//2))/(1+tan^(2)(x//2))+(1-tan^(2)(x//2))/(1+tan^(2)(x//2))])`
`=(pi)/(4)int_(0)^(pi//2)(sec^(2)(x//2))/(1-tan^(2)(x//2)+2tan(x//2))dx`
यदि `tan.(x)/(2)=t rArr sec^(2).(x)/(2)dx=2dt`
तथा सीमाएँ t = 0 पर वx = 0 पर `x=pi//2,` तब
`=(pi)/(4)int_(0)^(i//2)(sec^(2)(x//2))/(1-tan^(2)(x//2)+2tan(x//2))dx`
`=(pi)/(4)int_(0)^(1)(2dt)/(1-t^(2)+2t)`
`=(pi)/(2)int_(0)^(1)(dt)/([(sqrt2)^(2)-(t-1)^(2)])`
`=(pi)/(2).(1)/(2sqrt2)log[(sqrt2+(t-1))/(sqrt2-(t-1))]_(0)^(1)`
`=(pi)/(4sqrt2)log[(sqrt2+1)/(sqrt2-1)]`