Integration via Pythagorian Identity

anti-derivatives * anti-differentiation * integration

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

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

Integrating Sin/Cos

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

\end{aligned} $$

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