Question

`int_0^(pi/2) sqrt(cos theta) * sin^2 theta "d" theta` = ______.

Options

• `(- 20)/21`

• `(- 8)/21`

• `20/21`

• `8/21`

Answer 1

`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`