integral of sine cubed times cosine cubed

Asked Jun 17 '16 at 04:12

Active Jun 17 '16 at 04:31

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For the $\int \sin^3 x\cos^3 x \,dx$ I got $\frac 16(cos x)^6 - \frac14cos^4 x + C$.

This seemed to work but my book had $\frac 14\sin^4 x - \frac 16 sin^6 x + C$.

This is similar to what I got but instead of cosine they had sine and they had the signs(+/-) switched. So my question is: are both correct?

Any help would be appreciated.

edited Jun 17 '16 at 04:31

user317176

asked Jun 17 '16 at 04:12

alex

Both are right. The rwo main expressions differ by a constant. – André Nicolas Jun 17 '16 at 04:17
Suggestion: Differentite both expressions and see what you get. – mickep Jun 17 '16 at 04:17
Maybe faster: notice that the integrand function is symmetric around $x=\frac{\pi}{2}$. – Jack D'Aurizio Jun 17 '16 at 04:19
Remember that if $f$ and $g$ are primitives of a function $F$ then $f-g$ is a constant function. – Ángel Mario Gallegos Jun 17 '16 at 04:20
If you want to avoid differentiating, expand $$\frac16(\cos^2x+\sin^2x)^3-\frac14(\cos^2x+\sin^2x)^2.$$ – Did Jun 17 '16 at 05:25

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