Solutions - 0040-06

(1) Change variable by substituting $y=7x$ and $dy=7dx$:

$$\begin{array}{c}\int \sin 7x{\sec }^{5}7xdx\gg \gg \frac{1}{7}\int \sin 7x{\sec }^{5}7x7dx\\ \\ \gg \gg \frac{1}{7}\int \sin y{\sec }^{5}ydy\end{array}$$

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(2) Identify $\sin y\sec y=\tan y$:

$$\gg \gg \frac{1}{7}\int \tan y{\sec }^{4}ydy$$

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(3) Perform $u$-sub with $u=\sec y$ and thus $du=\sec y\tan ydy$:

$$\begin{array}{c}\gg \gg \frac{1}{7}\int {\sec }^{3}y(\sec y\tan y)dy\gg \gg \frac{1}{7}\int u^{3}du\\ \\ \gg \gg \frac{u^4}{28}+C\gg \gg \frac{{\sec }^{4}y}{28}+C\gg \gg \frac{{\sec }^4(7x)}{28}+C\end{array}$$