The Notes: `int_0^(pi/4) sec^4(theta) tan^4(theta) d theta` Evaluate the integral

Thursday, November 5, 2009

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{4}}}}}{{\sec}}^{{4}}{\left(\theta\right)}{{\tan}}^{{4}}{\left(\theta\right)}{d}\theta$ Evaluate the integral

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{4}}}}}{{\sec}}^{{4}}{\left(\theta\right)}{{\tan}}^{{4}}{\left(\theta\right)}{d}\theta$

Let's evaluate the indefinite integral by rewriting the integrand as,

$\displaystyle\int{{\sec}}^{{3}}{\left(\theta\right)}{{\tan}}^{{4}}{\left(\theta\right)}{d}\theta=\int{{\sec}}^{{2}}{\left(\theta\right)}{{\sec}}^{{2}}{\left(\theta\right)}{{\tan}}^{{4}}{\left(\theta\right)}{d}\theta$

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

$\displaystyle=\int{\left({1}+{{\tan}}^{{2}}{\left(\theta\right)}\right)}{{\sec}}^{{2}}{\left(\theta\right)}{{\tan}}^{{4}}{\left(\theta\right)}{d}\theta$

Now apply integral substitution,

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

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

$\displaystyle=\int{\left({1}+{{u}}^{{2}}\right)}{{u}}^{{4}}{d}{u}$

$\displaystyle=\int{\left({{u}}^{{4}}+{{u}}^{{6}}\right)}{d}{u}$

$\displaystyle=\int{{u}}^{{4}}{d}{u}+\int{{u}}^{{6}}{d}{u}$

$\displaystyle=\frac{{{u}}^{{5}}}{{5}}+\frac{{{u}}^{{7}}}{{7}}$

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

$\displaystyle=\frac{{1}}{{5}}{{\tan}}^{{5}}{\left(\theta\right)}+\frac{{1}}{{7}}{{\tan}}^{{7}}{\left(\theta\right)}$

Add a constant to the solution,

$\displaystyle=\frac{{1}}{{5}}{{\tan}}^{{5}}{\left(\theta\right)}+\frac{{1}}{{7}}{{\tan}}^{{7}}{\left(\theta\right)}+{C}$

Now let's evaluate the definite integral,

$\displaystyle{\int_{{0}}^{{\frac{\pi}{{4}}}}}{{\sec}}^{{4}}{\left(\theta\right)}{{\tan}}^{{4}}{\left(\theta\right)}{d}\theta={{\left[\frac{{1}}{{5}}{{\tan}}^{{5}}{\left(\theta\right)}+\frac{{1}}{{7}}{{\tan}}^{{7}}{\left(\theta\right)}\right]}_{{0}}^{{\frac{\pi}{{4}}}}}$

$\displaystyle={\left[\frac{{1}}{{5}}{{\tan}}^{{5}}{\left(\frac{\pi}{{4}}\right)}+\frac{{1}}{{7}}{{\tan}}^{{7}}{\left(\frac{\pi}{{4}}\right)}\right]}-{\left[\frac{{1}}{{5}}{{\tan}}^{{5}}{\left({0}\right)}+\frac{{1}}{{7}}{{\tan}}^{{7}}{\left({0}\right)}\right]}$