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

Explanation

माना `I=int_(0)^(pi//2)(sin^(2)x)/(sinx+cosx)dx" …(i)"`
`rArr" "I=int_(0)^(pi//2)(sin^(2)((pi)/(2)-x))/(sin((pi)/(2)-x)+cos((pi)/(2)-x))dx`
[प्रगुण (4 ) से ]
`" "=int_(0)^(pi//2)(cos^(2)x)/(cosx+sinx)dx" ...(2)"`
समीकरण (1 ) और (2 ) को जोड़ने पर
`2I=int_(0)^(pi//2)(sin^(2)x+cos^(2)x)/(cosx+sinx)dx`
`=int_(0)^(pi//2)(1)/(cosx+sinx)dx`
`=int_(0)^(pi//2)(1)/(sqrt2((1)/(sqrt2)cosx+(1)/(sqrt2)sinx))dx`
`=(1)/(sqrt2)int_(0)^(pi//2)(1)/(cos.(pi)/(4)cosx+sin.(pi)/(4)sinx)dx`
`=(1)/(sqrt2)int_(0)^(pi//2)(1)/(cos(x-(pi)/(4)))dx`
`=(1)/(sqrt2)int_(0)^(pi//2)sec(x-(pi)/(4))dx`
`=(1)/(sqrt2)[log|sec(x-(pi)/(4))+tan(x-(pi)/(4))|]_(0)^(pi//2)`
`=(1)/(sqrt2)[log|sec.(pi)/(4)+tan.(pi)/(4)|-log|sec(-(pi)/(4))+tan(-(pi)/(4))|]`
`=(1)/(sqrt2)[log(sqrt2+1)-log(sqrt2-1)]`
`=(1)/(sqrt2)log|(sqrt2+1)/(sqrt2-1)|`
`rArr" "I=(1)/(2sqrt2)log(sqrt2+1)^(2)`
`" "=(2)/(2Sqrt2)log(sqrt2+1)`
`" "=(1)/(sqrt2)log(sqrt2+1)` यही सिद्ध करना था ।