The Notes: `int tan^2(theta) sec^4(theta) d theta` Evaluate the integral

$\displaystyle\int{{\tan}}^{{2}}{\left(\theta\right)}{{\sec}}^{{4}}{\left(\theta\right)}{d}\theta$

Rewrite the integrand as,

$\displaystyle=\int{{\sec}}^{{2}}{\left(\theta\right)}{{\sec}}^{{2}}{\left(\theta\right)}{{\tan}}^{{2}}{\left(\theta\right)}{d}\theta$

Now use the identity: $\displaystyle{{\sec}}^{{2}}{\left({x}\right)}={1}+{{\tan}}^{{2}}{\left({x}\right)}$

$\displaystyle=\int{\left({1}+{{\tan}}^{{2}}{\left(\theta\right)}\right)}{{\sec}}^{{2}}{\left(\theta\right)}{{\tan}}^{{2}}{\left(\theta\right)}{d}\theta$

Now apply integral substitution,

Let $\displaystyle{u}={\tan{{\left(\theta\right)}}}$

$\displaystyle{d}{u}={{\sec}}^{{2}}{\left(\theta\right)}{d}\theta$

$\displaystyle=\int{\left({1}+{{u}}^{{2}}\right)}{{u}}^{{2}}{d}{u}$

$\displaystyle=\int{\left({{u}}^{{2}}+{{u}}^{{4}}\right)}{d}{u}$

$\displaystyle=\frac{{{u}}^{{3}}}{{3}}+\frac{{{u}}^{{5}}}{{5}}$

Substitute back $\displaystyle{u}={\tan{{\left(\theta\right)}}}$

$\displaystyle=\frac{{{{\tan}}^{{3}}{\left(\theta\right)}}}{{3}}+\frac{{{{\tan}}^{{5}}{\left(\theta\right)}}}{{5}}$

add a constant C to the solution,

$\displaystyle=\frac{{{{\tan}}^{{3}}{\left(\theta\right)}}}{{3}}+\frac{{{{\tan}}^{{5}}{\left(\theta\right)}}}{{5}}+{C}$

The Notes

Monday, October 10, 2016

$\displaystyle\int{{\tan}}^{{2}}{\left(\theta\right)}{{\sec}}^{{4}}{\left(\theta\right)}{d}\theta$ Evaluate the integral

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