Question

how do you integrate
sin (3pi/2x + pi/4)dx

Answer 1

∫sin (3pi/2x + pi/4)dx*

Answer 2

just notice
d/dx(cos (3pi/2x + pi/4))=-3pi/2 *sin (3pi/2x + pi/4)

Answer 3

...so\[u=\cos(\frac{3\pi}2x+\frac\pi4)\implies du=-\frac{3\pi}2\sin(\frac{3\pi}2x+\frac\pi4)dx\]\[-\frac2{3\pi}du\implies\sin(\frac{3\pi}2x+\frac\pi4)dx\]