Question

Evaluate the integral by using substitution.

`int_0^(pi/2) sqrt(sin phi) cos^5 phidphi`

Answer 1

Let  `I = int_0^(pi/2) sqrtsin phi cos^5 phi  d  phi`

`int_0^(pi/2) sin^(1/2) phi cos^4 phi cos phi   d  phi`

`int_0^(pi/2) sin^(1/2) phi. (1 - sin^2 phi)^2 . cos phi  d  phi`

On substituting `sin phi = t`,

`cos phi  d  phi = dt` and `phi = 0, t = 0,` When `phi = pi/2 t = 1`

Hence, `I = int_0^1  t^(1/2) (1 - t^2)^2 dt`

`I = int_0^1  t^(1/2) (1 + t^4 - 2t^2) dt`

`= int_0^1 (t^(1/2) + t^(9/2) - 2t^(5/2)) dt`

`= 2/3 [t^3]_0^1 + 2/11 [t^(11/2)]_0^1 - 2 xx 2/7 [t^(7/2)]_0^1`

`= 2/3 + 2/11 - 4/7`

`= (154 + 42 - 132)/231`

`= 64/231`