Question

evaluate the integral I= integral between 0 and pi/2 of cosx{2f'(sinx)-3) dx when f(0)=3 and f(1)=5

Answer 1

put sin(x)=t,
cos(x)dx=dt

I= \[\int\limits_{0}^{1} (2f'(t)-3) dt\]
\[0[2f(t) - 3t]1\]
=7-6
=1

Answer 2

ohhhh. Thank you:)

Answer 3

i dont understand the question here