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प्रश्न
`int_0^(pi/2) sqrt(cos theta) * sin^2 theta "d" theta` = ______.
पर्याय
`(- 20)/21`
`(- 8)/21`
`20/21`
`8/21`
MCQ
रिकाम्या जागा भरा
उत्तर
`int_0^(pi/2) sqrt(cos theta) * sin^2 theta "d" theta` = `underline(8/21)`.
Explanation:
Let I = `int_0^(pi/2) sqrt(cos theta) * sin^2 theta "d" theta`
= `int_0^(pi/2) sqrt(cos theta) * sin theta (1 - cos^2theta)"d"theta`
Put cos θ = t
⇒ - sin θ dθ = dt
⇒ sin θ dθ = - dt
If θ = 0, t = 1 and θ = `pi/2`, t = 0
`therefore "I" = int_1^0 sqrt"t" (1 - "t"^2)(- "dt")`
`= int_1^0 ("t"^(1//2) - "t"^(5//2))`dt
`= ("t"^(3//2)/(3/2) - "t"^(7//2)/(7/2))_0^1`
`= (2/3 "t"^(3//2) - 2/7 "t"^(7//2))_0^1`
`= (2/3 - 2/7) - (0 - 0)`
`= (14 - 6)/21 = 8/21`
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