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OpenStudy (anonymous):
how do you integrate sin (3pi/2x + pi/4)dx
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OpenStudy (anonymous):
∫sin (3pi/2x + pi/4)dx*
OpenStudy (anonymous):
just notice d/dx(cos (3pi/2x + pi/4))=-3pi/2 *sin (3pi/2x + pi/4)
OpenStudy (turingtest):
...so\[u=\cos(\frac{3\pi}2x+\frac\pi4)\implies du=-\frac{3\pi}2\sin(\frac{3\pi}2x+\frac\pi4)dx\]\[-\frac2{3\pi}du\implies\sin(\frac{3\pi}2x+\frac\pi4)dx\]
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