Cynthia Bell

2022-01-04

Solving limits:

$f\left(x\right)=\underset{n\to \mathrm{\infty}}{lim}\left(n{\int}_{0}^{\frac{\pi}{4}}{\left(\mathrm{tan}x\right)}^{n}dx\right)$

Hector Roberts

Beginner2022-01-05Added 31 answers

You may perform the change of variable $u=\mathrm{tan}x$ to get easily

${I}_{n}{\textstyle \phantom{\rule{0.222em}{0ex}}}={\int}_{0}^{\frac{\pi}{4}}{\left(\mathrm{tan}x\right)}^{n}dx={\int}_{0}^{1}\frac{{u}^{n}}{1+{u}^{2}}du$

Then you may just integrate by parts,

${I}_{n}={\int}_{0}^{1}\frac{{u}^{n}}{1+{u}^{2}}du=\frac{{u}^{n+1}}{(n+1)}\frac{1}{1+{u}^{2}}{\mid}_{0}^{1}+\frac{2}{(n+1)}{\int}_{0}^{1}\frac{{u}^{n+2}}{{(1+{u}^{2})}^{2}}du$

$=\frac{1}{2}\frac{1}{(n+1)}+\frac{2}{(n+1)}{\int}_{0}^{1}\frac{{u}^{n+2}}{{(1+{u}^{2})}^{2}}du$ (1)

Observing that

$0\le {\int}_{0}^{1}\frac{{u}^{n+2}}{{(1+{u}^{2})}^{2}}du\le {\int}_{0}^{1}{u}^{n}du=\frac{1}{n+1}$

gives

$0\le \frac{2}{(n+1)}{\int}_{0}^{1}\frac{{u}^{n}}{{(1+{u}^{2})}^{2}}du\le \frac{2}{{(n+1)}^{2}}$ (2)

Then combining (1) and (2) leads to

$\underset{n\to +\mathrm{\infty}}{lim}n{\int}_{0}^{\frac{\pi}{4}}{\left(\mathrm{tan}x\right)}^{n}dx=\underset{n\to +\mathrm{\infty}}{lim}n{I}_{n}=\frac{12}{}$

Then you may just integrate by parts,

Observing that

gives

Then combining (1) and (2) leads to

Orlando Paz

Beginner2022-01-06Added 42 answers

Heres

karton

Expert2022-01-11Added 613 answers

After letting

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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