Solutions - 0040-05

(a) Select $u=\tan x$ and thus $du={\sec }^{2}xdx$:

$$\begin{array}{cc}\int {\tan }^{3}x{\sec }^{2}xdx& \gg \gg \int u^{3}du\\ & \\ & \gg \gg \frac{u^4}{4}+C\\ & \\ & \gg \gg \frac{{\tan }^{4}x}{4}+C\end{array}$$

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(b)

(1) Select $u=\sec x$ and thus $du=\sec x\tan xdx$:

$$\int {\tan }^{3}x{\sec }^{2}xdx\gg \gg \int {\tan }^{2}x\sec x(\sec x\tan x)dx$$

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(3) Swap even bunch using $1+{\tan }^{2}x={\sec }^{2}x$:

$$\gg \gg \int ({\sec }^{2}x-1)\sec x(\sec x\tan x)dx$$

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(4) Perform $u$-sub with $u=\sec x$ and integrate:

$$\begin{array}{cc}& \gg \gg \int (u^2-1)udu\\ & \\ & \gg \gg \frac{u^4}{4}-\frac{u^2}{2}+C\\ & \\ & \gg \gg \frac{{\sec }^{4}x}{4}-\frac{{\sec }^{2}x}{2}+C\end{array}$$