Question

निम्नलिखित का मान निकालिए-

`int_(pi/3)^(pi/2) sqrt(1 + cosx)/(1 - cos x)^(5/2)  "d"x`

Answer 1

मान लीजिए 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`

`sin  x/2` = t रखिए

⇒ `1/2 cos  x/2 "d"x` = dt

⇒ `cos  x/2 "d"x` = 2dt

सीमाएँ बदलना, हमारे पास है

जब x = `pi/3`

`sin  pi/6` = t

∴ t = `1/2`

जब 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`

अत:, I = `3/2`.