`int_(pi//3)^(pi//2) (sqrt(1+cosx))/((1-cosx)^(5//2))\ dx`

Answer

N/a

Explanation

Let `I = int_(pi//3)^(pi//2) (sqrt(1+cosx))/((1-cosx)^(5//2))dx`
`= int_(pi//3)^(pi//2) (sqrt(1+cosx))/((1+cosx)^(2)sqrt(1+cosx))dx`
`int_(pi3)^(pi2)(1)/((1-cos^(2)x))dx = int_(pi3)^(pi2)(1)/(sin^(2)x)dx`
`= int_(pi3)^(pi2)cosec^(2)xdx = [-cotx]_(pi//3)^(pi//2)`
`= -[cot'(pi)/(2) - cot'(pi)/(3)] = - [0 - (1)/(sqrt(3))] = + 1/(sqrt(3))`
Alternate Method
Let ` I = int_(pi//2)^(pi//2) (sqrt(1+cosx))/((1-cosx)^(5//2)) dx = int_(pi//3)^(pi//2)((2cos^(2)'(x)/(2))^(1//2))/((2sin^(2)'(x)/(2))^(5//2))dx`
` = (sqrt(2))/(4sqrt(2))int_(pi//3)^(pi//2) (cos(x/2))/(sin^(5)(x/2)) dx = 1/4int_(pi//3)^(pi//2)(cos(x/2))/(sin^(5)(x/2)) dx`
Put `sin'x/2 = t`
`rArr cos'x/(2).(1)/(2) dx = dt`
`rArr cos'x/2 dx = 2dt`
As `x rarr pi/3`, then `t rarr 1/3`
and `x rarr pi/2`, then `t rarr 1/(sqrt(2))`
`:. I = 2/4 int_(1//2)^(1sqrt(2))(dt)/(t^(5)) = (1)/(2)[(t^(-5+1))/(-5+1)]_(1//2)^(1//sqrt(2))`
`= -1/8 [(1)/(((1)/(sqrt(2)))^(4) - ((1)/((1)/2))^(4))]`
`= - 1/8 (4-16) = 12/8 = 3/2`