91Ó°ÊÓ

The given integral is$∫ sec{ }^3 x ta{n}^{2}xdx$

• Use the trigonometric identities to rewrite the given integral.

$\begin{array}{rcl}∫ sec{ }^3 x ta{n}^{2}xdx& =& ∫{\sec }^3\left(x\right)\left(-1+{\sec }^2\left(x\right)\right)dx\\ & =& ∫ -{\sec }^3\left(x\right)+{\sec }^5\left(x\right)dx\\ & =& ∫ -{\sec }^3\left(x\right)dx+∫{\sec }^5\left(x\right)dx\end{array}$

• Integrate the integral $\begin{array}{rcl}& & ∫{\sec }^5\left(x\right)dx\end{array}$.

role="math" localid="1649014338432" $\begin{array}{rcl}∫{\sec }^5\left(x\right)dx& =& ∫{\sec }^3\left(x\right){\sec }^2\left(x\right)dx\\ & =& {\sec }^3\left(x\right)\tan \left(x\right)-3∫{\sec }^2\left(x\right)\sec \left(x\right)\tan \left(x\right)\tan \left(x\right)dx\\ & =& {\sec }^3\left(x\right)\tan \left(x\right)-3∫{\sec }^3\left(x\right){\tan }^2\left(x\right)dx\\ & =& {\sec }^3\left(x\right)\tan \left(x\right)-3∫{\sec }^5\left(x\right)dx+3∫{\sec }^3\left(x\right)dx\end{array}$

• This implies that

role="math" localid="1649014573251" $\begin{array}{rcl}4∫{\sec }^5\left(x\right)dx& =& {\sec }^3\left(x\right)\tan \left(x\right)+3∫{\sec }^3\left(x\right)dx\\ ∫{\sec }^5\left(x\right)dx& =& \frac{1}{4}{\sec }^3\left(x\right)\tan \left(x\right)+\frac{3}{4}∫{\sec }^3\left(x\right)dx\\ ∫{\sec }^5\left(x\right)dx-∫{\sec }^3\left(x\right)dx& =& \frac{1}{4}{\sec }^3\left(x\right)\tan \left(x\right)-\frac{1}{4}∫{\sec }^3\left(x\right)dx\end{array}$

• Integrate the integral $\begin{array}{rcl}& & ∫{\sec }^3\left(x\right)dx\end{array}$.

role="math" localid="1649014450741" $\begin{array}{rcl}∫{\sec }^3\left(x\right)dx& =& ∫\sec \left(x\right){\sec }^2\left(x\right)dx\\ & =& \sec \left(x\right)\tan \left(x\right)-∫\sec \left(x\right)\tan \left(x\right)\tan \left(x\right)dx\\ & =& \sec \left(x\right)\tan \left(x\right)-∫\sec \left(x\right){\tan }^2\left(x\right)dx\\ & =& \sec \left(x\right)\tan \left(x\right)-∫{\sec }^3\left(x\right)dx+∫\sec \left(x\right)dx\\ & =& \sec \left(x\right)\tan \left(x\right)+\log \left|\sec \left(x\right)+\tan \left(x\right)\right|-∫{\sec }^3\left(x\right)dx\end{array}$

• So,role="math" localid="1649014593933" $\begin{array}{rcl}2∫{\sec }^3\left(x\right)dx& =& \sec \left(x\right)\tan \left(x\right)+\log \left|\sec \left(x\right)+\tan \left(x\right)\right|\\ ∫{\sec }^3\left(x\right)dx& =& \frac{1}{2}\sec \left(x\right)\tan \left(x\right)+\frac{1}{2}\log \left|\sec \left(x\right)+\tan \left(x\right)\right|\end{array}$

• Substitute the value of $\begin{array}{rcl}& & ∫{\sec }^3\left(x\right)dx\end{array}$into $\begin{array}{rcl}∫{\sec }^5\left(x\right)dx-∫{\sec }^3\left(x\right)dx& =& \frac{1}{4}{\sec }^3\left(x\right)\tan \left(x\right)-\frac{1}{4}∫{\sec }^3\left(x\right)dx\end{array}$.

$\begin{array}{rcl}∫{\sec }^5\left(x\right)dx-∫{\sec }^3\left(x\right)dx& =& \frac{1}{4}{\sec }^3\left(x\right)\tan \left(x\right)-\frac{1}{4}\left(\frac{1}{2}\sec \left(x\right)\tan \left(x\right)+\frac{1}{2}\log \left|\sec \left(x\right)+\tan \left(x\right)\right|\right)\end{array}$