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Question
Evaluate the following:
`int_(pi/3)^(pi/2) sqrt(1 + cosx)/(1 - cos x)^(5/2) "d"x`
Solution
Let I = `int_(pi/3)^(pi/2) sqrt(1 + cosx)/(1 - cos x)^(5/2) "d"x`
= `int_(pi/3)^(pi/2) sqrt(2cos^2 x/2)/(2sin^2 x/2)^(5/2) "d"x`
= `int_(pi/3)^(pi/2) (sqrt(2) cos x/2)/((2)^(5/2) sin^5 x/2) "d"x`
= `1/4 int_(pi/3)^(pi/2) (cos x/2)/(sin^5 x /2) "d"x`
Put `sin x/2` = t
⇒ `1/2 cos x/2 "d"x` = dt
⇒ `cos x/2 "d"x` = 2dt
Changing the limits, we have
When x = `pi/3`
`sin pi/6` = t
∴ t = `1/2`
When x = `pi/2`
`sin pi/4` = t
∴ t = `1/sqrt(2)`
∴ I = `1/4 xx 2 int_(1/2)^(1/sqrt(2)) "dt"/"t"^5`
= `1/2 xx (- 1/4) ["t"^-4]_(1/2)^(1/sqrt(2))`
= `- 1/8 [1/"t"^4]_(1/2)^(1/sqrt(2))`
= ` 1/8 [1/((1/sqrt(2))^4 - (1/(1/2))^4)]`
= `- 1/8 [4 - 16]`
= `- 1/8 xx (-12)`
= `3/2`
Hence, I = `3/2`.
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