Find f(pi/2) if f(pi)=1/2 and f’(x)=sinxcosx

Answer 1

f'(x) = sinxcosx = (1/2)sin(2x)

So, f(x) = ∫f'(x)dx = -(1/4)cos(2x) + C

Since f(π) = 1/2, we have -(1/4)cos(2π) + C = 1/2. So, -(1/4) + C = 1/2. Therefore, C = 3/4.

f(x) = -(1/4)cos(2x) + 3/4

f(π/2) = -(1/4)cosπ + 3/4 = 1

Answer 2

integral [sin cos]

U = sin dV = cos

dU = cos v = sin

sin^2 - integral [ cos sin]

so 2 * integral [ sin cos ] = sin ^2

the integral is (1/2) sin^2 + C

f(pi) = 1/2

(1/2) (sin pi)^2 + c = 1/2

(1/2) * 0 + c = 1/2

c = 1/2

f(x) = (1/2) sin^2 + 1/2

f(pi/2) = (1/2) [sin(pi/2)]^2 + 1/2

= (1/2) 1^2 + 1/2

= 1