ज्ञात करें `intsqrt(1+sin theta)d theta=intsqrt(1+cos(pi/2-theta))d theta`

Explanation

`=intsqrt(2cos^2((pi//2-theta)/2))d theta" "[:." "1+cos x = 2 cos^2""x/2]`
`=sqrt2intcos(pi/4-theta/2)d theta`
`=sqrt2*(sin(pi//4-theta//2))/(-1/2)+c=-2sqrt2sin(pi/4-theta/2)+c`
`=2sqrt2sin(theta/2-theta/4)+c" "[ becausesin(-theta)=-sintheta]`
`sin(theta/2-pi/4)=""theta/2cos""pi/4-cos""theta/2=1/sqrt2(sin""theta/2-cos""theta/2)`
`:." "I=2(sin""theta/2-cos""theta/2)+c`