#### Question 9.3.1.

We already know we can do the following problem using u-substitution, with \(u = \sin(8 x )\)

\(\displaystyle \int \sin^{6}\!\left(8x\right)\cos\!\left(8x\right) \,dx \ = \ \)

We see that \(\displaystyle \int \sin^{6} (8 x) \cos^3(8 x) \,dx \ \) is solved similarly, because

- the exponent of \(\sin\) is odd.
- the exponent of \(\cos\) is odd.
- None of the above

So we’ll still use u-substitution with \(u = \sin(8 x )\text{,}\) after first rewriting \(\cos^3 (8 x) = \cos^2(8 x) \cos(8 x )\) and using a trig identity.

Select the trigonometric identity needed to solve this problem.

- \(\displaystyle \displaystyle \sin^2 u + \cos^2 u = 1\)
- \(\displaystyle \displaystyle \cos^2 u = \frac{1+\cos(2 u)}{2}\)
- \(\displaystyle \displaystyle \sin^2 u = \frac{1-\cos(2 u)}{2}\)
- \(\displaystyle \displaystyle \tan^2 u +1 = \sec^2 u\)
- \(\displaystyle \displaystyle 1+ \cot^2 u = \csc^2 u\)
- None of the above

Finally, put it all together to calculate \(\displaystyle \int \sin^{6}\!\left(8x\right)\cos^{3}\!\left(8x\right) \,dx \ = \ \)

## Hint.

The integral involves \(\sin u\) and \(\cos u\) where the exponent of \(\cos u\) is an odd, positive integer, namely, 3. So we factor out a copy of \(\cos(8 x)\) and use the identity \(\sin^2 u + \cos^2 u = 1\) to turn the remaining \(\cos^2(8 x)\) into an expression involving \(\sin^2(8 x)\text{.}\)

Thus the integral can be rewritten as

\begin{equation*}
\int \sin^{6}\!\left(8x\right)\cos^{3}\!\left(8x\right) \, dx = \int \sin^{6} (8 x) \cos^2(8 x) \cos(8 x) \, dx
\end{equation*}

\begin{equation*}
= \int \sin^{6} (8 x) \bigg(1-\sin^2(8 x)\bigg) \cos(8 x) \, dx
\end{equation*}

and we use u-substitution.

Letting \(u= \sin (8 x)\) so \(du = 8 \cos (8 x) \, dx\) and thus \(\displaystyle dx = \frac{ du}{8 \cos (8 x)}\text{.}\)

This turns the integral into

\begin{equation*}
\int u^{6} (1-u^2) \cos(8 x) \, \frac{ du}{8 \cos (8 x)} = \frac{1}{8}\int u^{6} (1-u^2) \, du
\end{equation*}

which can be multiplied out, and then each part integrated by power rule.