Question $5$ of $15$

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Evaluate the integral.

$($Use symbolic notation and fractions where needed. Use C for the arbitrary constant.$)$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}co{t}^3(6x)dx=\frac{1}{2}co{t}^2(6x)+\frac{1}{6}logsin(6x)+C$

Logarithin function is missing parcetheses around its argument

Sorry, that's incorrect.

Begin by rewriting the integral in terms of sine and cosine and factoring the integrand using the identity ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}co{s}^n(x)dx={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}co{s}^{n-1}(x)cos(x)dx,$ where $n3$ is an odd integer.

Next, rewrite the integrand using the identity $si{n}^2(x)+co{s}^2(x)=1.$

Then use the substitution $u=sin(6x),du=6cos(6x)dx.$

$+100$ points still possible

Question Source: Rogreski $4$e