How do you integrate #(sinx)(cosx)(cos2x)dx#?

2 Answers
Mar 15, 2015

First use a double-angle formula to replace #cos(2x)# by #2cos^{2}(x)-1#. Then distribute #cos(x)# through to rewrite your integrand as #(2cos^{3}(x)-cos(x))sin(x)#. Now do a substitution: #u=cos(x), du=-sin(x)dx#, making your integral transform to #\int(u-2u^{3})du=u^{2}/2-u^{4}/2+C=\frac{1}{2}\cos^{2}(x)-\frac{1}{2}\cos^{4}(x)+C.# There are lots of alternative ways of writing this answer because of all the trigonometric identities out there. You could check, for instance, that it is equivalent to #-\frac{1}{16}\cos(4x)+C#.

Mar 15, 2015

You can try this:
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where at the end you treat #sin# and #cos# as #x# in a normal integral where you would use the form: #x^(n+1)/(n+1)# to integrate #x^n#.