Integration via Pythagorian Identity

integration

$si{n}^2(x)+co{s}^2(x)=1$

$ta{n}^2(x)+1=se{c}^2(x)$

Integrating Sin/Cos

$$\begin{array}{rl}& \int si{n}^4(x)co{s}^3(x)dx\\ & \\ & =\int si{n}^4(x)\cdot [1-si{n}^2(x)]cos(x)dx\\ & \\ u& =sin(x)\\ & \\ du& =cos(x)dx\\ & \\ & =\int u^4(1-u^2)du\\ & \\ & =\int u^4-u^{6}du\\ & \\ & =\frac{1}{5}{u}^5-\frac{1}{7}{u}^7+C\\ & \\ & =\frac{1}{5}sin(x{)}^5-\frac{1}{7}sin(x{0}^7+C\\ & \end{array}$$ ​

Integration Tan/Sec

Example

$$\begin{array}{rl}& \int {\tan }^2(x){\sec }^4(x)dx\\ & \\ u& =\tan (x)\\ & \\ du& ={\sec }^2(x)dx\\ & \\ & \int {\tan }^2(x)[{\tan }^2(x)+1]{\sec }^2(x)du\\ & \\ & \int u^2[u^2+1]du\\ & \\ & \int u^4+u^{2}du\\ & \\ & =\frac{1}{5}{u}^5+\frac{1}{3}{u}^3+C\\ & \\ & =\frac{1}{5}ta{n}^5(x)+\frac{1}{3}ta{n}^3(x)+C\end{array}$$

Example 2

$$\begin{array}{rl}\int ta{n}^3(x)se{c}^3(x)dx& \\ & \\ u& =sec(x)\\ & \\ du& =tan(x)sec(x)dx\\ & \\ \int [se{c}^2(x)-1]tan(x)se{c}^2(x)sec(x)dx& \\ & \\ \int u^2[u^2-1]du& \\ & \\ \int u^4-u^{2}du& \\ & \\ \frac{1}{5}{u}^5-\frac{1}{3}{u}^3+C& \\ & \\ \frac{1}{5}sec(x{)}^5-\frac{1}{3}sec(x{)}^3+C& \\ & \end{array}$$