समाकलनो का मान ज्ञात कीजिएः
`int_(0)^(pi//2) (1)/(1+sinx)dx`

Explanation

यहाँ `I=underset(0)overset(pi//2)int (1)/(1+sin x)dx`
`=underset(0)overset(pi//2)int ((1)/(1+sin x)xx(1-sin x)/(1-sin x))dx`
`=underset(0)overset(pi//2)int (1-sin x)/(1-sin^(2)x)dx=underset(0)overset(pi//2)int (1- sin )/(cos^(2)x)dx`
`=underset(0)overset(pi//2)int ((1)/(cos^(2)x)-(sin x)/(cos^(2)x))dx`
`=underset(0)overset(pi//2)int (sec^(2)x-tan xsec x)dx`
`=[tan x-sec x]_(0)^(pi//2)`
`=underset(x to pi/2)lim (tan x-sec x)-(tan 0-sec 0)`
`=underset(x to pi/2)lim (sin x-1)/(cos x)-(0-1)`
`=underset(x to pi/2)lim ((cos x)/(-sin x))+1`
=0+1=1