to integrate 3cos(4x-3)dx in definate integral form with limi 0 to 1

Question

to integrate 3cos(4x-3)dx in definate integral form with limit 0 to 1

Answer 1

Given ∫3cos(4x - 3)dx with limits 0→1

3∫cos(4x-3)dx

let ...... u=4x-3

du/dx=4.......... dx=du/4

=3∫cosu du/4

3/4∫cosu

={3/4sin(4x-3)} with the limits 1 and 0

={3/4sin(4 - 3)} - {3/4sin(0 - 3)}

=3/4sin1

=3/4(pi/2)