The Notes: `int_(pi/6)^(pi/3) csc^3(dx)` Evaluate the integral

Thursday, March 19, 2015

$\displaystyle{\int_{{\frac{\pi}{{6}}}}^{{\frac{\pi}{{3}}}}}{{\csc}}^{{3}}{\left({\left.{d}{x}\right.}\right)}$ Evaluate the integral

$\displaystyle{\int_{{\frac{\pi}{{6}}}}^{{\frac{\pi}{{3}}}}}{{\csc}}^{{3}}{\left({x}\right)}{\left.{d}{x}\right.}$

Apply the integral substitution,

Let $\displaystyle{u}={\tan{{\left(\frac{{x}}{{2}}\right)}}}$

$\displaystyle\Rightarrow{d}{u}=\frac{{1}}{{2}}{{\sec}}^{{2}}{\left(\frac{{x}}{{2}}\right)}{\left.{d}{x}\right.}$

using pythagorean identity: $\displaystyle{1}+{{\tan}}^{{2}}{\left(\theta\right)}={{\sec}}^{{2}}{\left(\theta\right)}$

$\displaystyle{d}{u}=\frac{{1}}{{2}}{\left({1}+{{\tan}}^{{2}}{\left(\frac{{x}}{{2}}\right)}\right)}{\left.{d}{x}\right.}$

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

$\displaystyle\Rightarrow{\left.{d}{x}\right.}={\left(\frac{{2}}{{{1}+{{u}}^{{2}}}}\right)}{d}{u}$

$\displaystyle{\csc{{\left({x}\right)}}}=\frac{{1}}{{\sin{{\left({x}\right)}}}}=\frac{{1}}{{{2}{\sin{{\left(\frac{{x}}{{2}}\right)}}}{\cos{{\left(\frac{{x}}{{2}}\right)}}}}}$

$\displaystyle=\frac{{\frac{{1}}{{{{\cos}}^{{2}}{\left(\frac{{x}}{{2}}\right)}}}}}{{\frac{{{2}{\sin{{\left(\frac{{x}}{{2}}\right)}}}{\cos{{\left(\frac{{x}}{{2}}\right)}}}}}{{{{\cos}}^{{2}}{\left(\frac{{x}}{{2}}\right)}}}}}$

$\displaystyle=\frac{{{{\sec}}^{{2}}{\left(\frac{{x}}{{2}}\right)}}}{{{2}{\tan{{\left(\frac{{x}}{{2}}\right)}}}}}$

$\displaystyle=\frac{{{1}+{{\tan}}^{{2}}{\left(\frac{{x}}{{2}}\right)}}}{{{2}{\tan{{\left(\frac{{x}}{{2}}\right)}}}}}$

$\displaystyle=\frac{{{1}+{{u}}^{{2}}}}{{{2}{u}}}$

Now let's evaluate the indefinite integral,

$\displaystyle\int{{\csc}}^{{3}}{\left({x}\right)}{\left.{d}{x}\right.}=\int{{\left(\frac{{{1}+{{u}}^{{2}}}}{{{2}{u}}}\right)}}^{{3}}{\left(\frac{{2}}{{{1}+{{u}}^{{2}}}}\right)}{d}{u}$

$\displaystyle=\int\frac{{{\left({1}+{{u}}^{{2}}\right)}}^{{2}}}{{{4}{{u}}^{{3}}}}{d}{u}$

$\displaystyle=\int\frac{{{1}+{2}{{u}}^{{2}}+{{u}}^{{4}}}}{{{4}{{u}}^{{3}}}}{d}{u}$

$\displaystyle=\frac{{1}}{{4}}\int{\left(\frac{{1}}{{{u}}^{{3}}}+\frac{{2}}{{u}}+{u}\right)}{d}{u}$

$\displaystyle=\frac{{1}}{{4}}{\left(\int\frac{{1}}{{{u}}^{{3}}}{d}{u}+{2}\int\frac{{1}}{{u}}{d}{u}+\int{u}{d}{u}\right)}$

$\displaystyle=\frac{{1}}{{4}}{\left({\left(\frac{{{u}}^{{-{3}+{1}}}}{{-{3}+{1}}}\right)}+{2}{\ln}{\left|{u}\right|}+\frac{{{u}}^{{2}}}{{2}}\right)}$

$\displaystyle=\frac{{1}}{{4}}{\left(-\frac{{1}}{{{2}{{u}}^{{2}}}}+{2}{\ln}{\left|{u}\right|}+\frac{{{u}}^{{2}}}{{2}}\right)}$

Substitute back $\displaystyle{u}={\tan{{\left(\frac{{x}}{{2}}\right)}}}$

$\displaystyle=\frac{{1}}{{4}}{\left(-\frac{{1}}{{{2}{{\tan}}^{{2}}{\left(\frac{{x}}{{2}}\right)}}}+{2}{\ln}{\left|{\tan{{\left(\frac{{x}}{{2}}\right)}}}\right|}+\frac{{1}}{{2}}{{\tan}}^{{2}}{\left(\frac{{x}}{{2}}\right)}\right.}$

$\displaystyle=\frac{{1}}{{4}}{\left(\frac{{1}}{{2}}{{\tan}}^{{2}}{\left(\frac{{x}}{{2}}\right)}-\frac{{1}}{{2}}{{\cot}}^{{2}}{\left(\frac{{x}}{{2}}\right)}+{2}{\ln}{\left|{\tan{{\left(\frac{{x}}{{2}}\right)}}}\right|}\right)}$

Add a constant C to the solution,

$\displaystyle{\int_{{\frac{\pi}{{6}}}}^{{\frac{\pi}{{3}}}}}{{\csc}}^{{3}}{\left({x}\right)}{\left.{d}{x}\right.}={{\left[\frac{{1}}{{4}}{\left(\frac{{1}}{{2}}{{\tan}}^{{2}}{\left(\frac{{x}}{{2}}\right)}-\frac{{1}}{{2}}{{\cot}}^{{2}}{\left(\frac{{x}}{{2}}\right)}+{2}{\ln}{\left|{\tan{{\left(\frac{{x}}{{2}}\right)}}}\right|}\right)}\right]}_{{\frac{\pi}{{6}}}}^{{\frac{\pi}{{3}}}}}$

$\displaystyle={\left[\frac{{1}}{{4}}{\left(\frac{{1}}{{2}}{{\tan}}^{{2}}{\left(\frac{\pi}{{6}}\right)}-\frac{{1}}{{2}}{{\cot}}^{{2}}{\left(\frac{\pi}{{6}}\right)}+{2}{\ln}{\left|{\tan{{\left(\frac{\pi}{{6}}\right)}}}\right|}\right]}-{\left[\frac{{1}}{{4}}{\left(\frac{{1}}{{2}}{{\tan}}^{{2}}{\left(\frac{\pi}{{12}}\right)}-\frac{{1}}{{2}}{{\cot}}^{{2}}{\left(\frac{\pi}{{12}}\right)}+{2}{\ln}{\left|{\tan{{\left(\frac{\pi}{{12}}\right)}}}\right|}\right]}\right.}\right.}$

$\displaystyle={\left[\frac{{1}}{{4}}{\left(\frac{{1}}{{2}}{{\left(\frac{{1}}{\sqrt{{{3}}}}\right)}}^{{2}}-\frac{{1}}{{2}}{{\left(\sqrt{{{3}}}\right)}}^{{2}}+{2}{\ln{{\left(\frac{{1}}{\sqrt{{{3}}}}\right)}}}\right]}-{\left[\frac{{1}}{{4}}{\left(\frac{{1}}{{2}}{{\left({2}-\sqrt{{{3}}}\right)}}^{{2}}-\frac{{1}}{{2}}{{\left({2}+\sqrt{{{3}}}\right)}}^{{2}}+{2}{\ln{{\left({2}-\sqrt{{{3}}}\right)}}}\right]}\right.}\right.}$

$\displaystyle={\left[\frac{{1}}{{4}}{\left(\frac{{1}}{{6}}-\frac{{3}}{{2}}-{\ln{{3}}}\right)}\right]}-{\left[\frac{{1}}{{4}}{\left(\frac{{1}}{{2}}{\left({4}-{4}\sqrt{{{3}}}+{3}\right)}-\frac{{1}}{{2}}{\left({4}+{4}\sqrt{{{3}}}+{3}+{2}{\ln{{\left({2}-\sqrt{{{3}}}\right)}}}\right)}\right]}\right.}$

$\displaystyle={\left[-\frac{{1}}{{3}}-\frac{{\ln{{\left({3}\right)}}}}{{4}}\right]}-{\left[\frac{{1}}{{4}}{\left(\frac{{7}}{{2}}-{2}\sqrt{{{3}}}-\frac{{7}}{{2}}-{2}\sqrt{{{3}}}+{2}{\ln{{\left({2}-\sqrt{{{3}}}\right)}}}\right]}\right.}$

$\displaystyle={\left[-\frac{{1}}{{3}}-\frac{{\ln{{\left({3}\right)}}}}{{4}}\right]}-{\left[\frac{{1}}{{4}}{\left(-{4}\sqrt{{{3}}}+{2}{\ln{{\left({2}-\sqrt{{{3}}}\right)}}}\right)}\right]}$

$\displaystyle={\left[-\frac{{1}}{{3}}-\frac{{\ln{{\left({3}\right)}}}}{{4}}\right]}-{\left[\frac{{1}}{{4}}{\left(-{4}\sqrt{{{3}}}+{\ln{{\left({4}+{3}-{4}\sqrt{{{3}}}\right)}}}\right)}\right]}$

$\displaystyle=-\frac{{1}}{{3}}-\frac{{\ln{{\left({3}\right)}}}}{{4}}+\sqrt{{{3}}}-\frac{{\ln{{\left({7}-{4}\sqrt{{{3}}}\right)}}}}{{4}}$

$\displaystyle=-\frac{{1}}{{3}}+\sqrt{{{3}}}-\frac{{\ln{{\left({3}\right)}}}}{{4}}-\frac{{\ln{{\left({7}-{4}\sqrt{{{3}}}\right)}}}}{{4}}$

$\displaystyle=-\frac{{1}}{{3}}+\sqrt{{{3}}}+\frac{{1}}{{4}}{\ln{{\left(\frac{{{7}+{4}\sqrt{{{3}}}}}{{3}}\right)}}}$

The Notes

Thursday, March 19, 2015

$\displaystyle{\int_{{\frac{\pi}{{6}}}}^{{\frac{\pi}{{3}}}}}{{\csc}}^{{3}}{\left({\left.{d}{x}\right.}\right)}$ Evaluate the integral

No comments:

Post a Comment

What are hearing tests?