Answer
$3sin\frac x 3 -sin^3 \frac x 3 +C$
Work Step by Step
$\int cos^3\frac x 3 dx$
$\int (cos^2 \frac x 3)(cos\frac x 3)dx$
$\int (cos\frac x 3-sin^2\frac x 3cos\frac x 3)dx$
$3\int cos \frac x 3 (\frac 1 3)dx-3\int sin^2\frac x 3(cos \frac x 3)(\frac 1 3)dx$
$u=\frac x 3, du=\frac 1 3 dx$
$3\int cosdu=3sinu+C=3sin\frac x 3 +C$
$u=sin\frac x 3, du=(cos\frac x 3)(\frac 1 3)dx$
$3\int u^2du=u^3+C=sin^3\frac x 3+C$
$3sin\frac x 3 -sin^3 \frac x 3 +C$
