Thread: trig integration help!

i am taking the integral of (sin(mx))^3 i.e. sin cubed of mx. i used broke it into sin squared mx times sin mx and then used substitution. i set u equal to cos(mx) and got -m^-1 du = sin (mx) dx

for the answer i got: -m^-1 * cos(mx) + (3*m)^-1 * cos^3(mx) + c
is this correct?

Originally Posted by pcgamer03

i am taking the integral of (sin(mx))^3 i.e. sin cubed of mx. i used broke it into sin squared mx times sin mx and then used substitution. i set u equal to cos(mx) and got -m^-1 du = sin (mx) dx

for the answer i got: -m^-1 * cos(mx) + (3*m)^-1 * cos^3(mx) + c
is this correct?

yes