`int(sinx)/sqrt(1+sinx)dx` का मान ज्ञात कीजिए।

Explanation

दिया है -` I = int(sinx)/(sqrt(1+sin x))dx = int((1+sin x) - 1)/(sqrt(1+sin x))`
` = int sqrt(1+sinx)dx - int(dx)/(sqrt(1+sin x))`
`= intsqrt(cos^(2).(x)/(2)+sin^(2)+(x)/(2)+2sin.(x)/(2)cos.(x)/(2)dx)`
`-int(dx)/(sqrt(cos^(2)(x//2)+sin^(2)(x//2)+2sin(x//2))cos(x//2))`
`= int[cos((x)/(2))+sin((x)/(2))]dx - int(dx)/([cos (x//2)+sin(x//2)])`
` = (2sin.(x)/(2) - cos .(x)/(2)) -(1)/(2).int(dx)/((1)/(sqrt(2)cos.(x)/(2)+(1)/(sqrt(2))sin.(x)/(2)))`
`=(2sin.(x)/(2)-2cos.(x)/(2)) - (1)/(sqrt(2)) . int(dx)/(sin((pi)/(2) +(pi)/(2)))`
` = (2sin.(x)/(2) - 2cos.(x)/(2)) -(1)/(sqrt(2)). intcosec((x)/(2) +(pi)/(4))dx`
` = 2 (sin.(x)/(2)-cos.(x)/(2)) - (1)/(sqrt(2)) xx 2 log[ tan ((x)/(4) +(pi)/(8))]+c`
` = 2(sin.(x)/(2) - cos.(x)/(2))- sqrt(2)[ tan ((pi)/(4) +(pi)/8)]+c`