Integrals

 ${\displaystyle \underset{0}{\overset{1}{\int }}1.{\left(1-x^n\right)}^{2n+1}dx}$ using by parts we get,

$\left(2n^2+n+1\right){\displaystyle \underset{0}{\overset{1}{\int }}{\left(1-x^n\right)}^{2n+1}dx=1177{\displaystyle \underset{0}{\overset{1}{\int }}{\left(1-x^n\right)}^{2n+1}dx}}$

$\Rightarrow 2n^2+n+1=1177\Rightarrow n=24or-\frac{49}{2}\therefore n=24$

 $tan\left(2ta{n}^{-1}\frac{1}{5}+se{c}^{-1}\frac{\sqrt[]{5}}{2}+2ta{n}^{-1}\frac{1}{8}\right)tan\left(2ta{n}^{-1}\frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5}\times \frac{1}{8}}+se{c}^{-1}\frac{\sqrt[]{5}}{2}\right)$

$=tan\left(ta{n}^{-1}\frac{3}{4}+ta{n}^{-1}\frac{1}{2}\right)=tan\left(ta{n}^{-1}\frac{\frac{3}{4}+\frac{1}{2}}{1-\frac{3}{8}}\right)$

$=tan\left(ta{n}^{-1}\frac{\frac{5}{4}}{\frac{5}{8}}\right)=2$

  ${\displaystyle \int \frac{\left(1-\frac{1}{\sqrt[]{3}}\right)\left(cosx-sinx\right)}{\left(1+\frac{2}{\sqrt[]{3}}sin2x\right)}}dx={\displaystyle \int \frac{\left(\frac{\sqrt[]{3}-1}{\sqrt[]{3}}\right)\sqrt[]{2}sin\left(\frac{\pi }{4}-x\right)}{\left(\frac{2}{\sqrt[]{3}}\right)\left(sin\frac{\pi }{3}+sin2x\right)}dx}$

$={\displaystyle \int \frac{\left(\frac{\sqrt[]{3}-1}{\sqrt[]{2}}\right)sin\left(\frac{\pi }{4}-x\right)}{\left(sin\frac{\pi }{3}+sin2x\right)}}dx={\displaystyle \int \frac{\left(\frac{\sqrt[]{3}-1}{2\sqrt[]{2}}\right)sin\left(\frac{\pi }{4}-x\right)}{sin\left(\frac{\pi }{6}+x\right)cos\left(\frac{\pi }{6}-x\right)}dx}$

$=\frac{1}{2}\left[lo{g}_e\left|tan\left(\frac{x}{2}+\frac{\pi }{12}\right)\right|-lo{g}_e\left|tan\left(\frac{x}{2}+\frac{\pi }{6}\right)\right|\right]+C=\frac{1}{2}lo{g}_e\left|\frac{tan\left(\frac{x}{2}+\frac{\pi }{12}\right)}{tan\left(\frac{x}{2}+\frac{\pi }{6}\right)}\right|+C$

 $a=\underset{n\to \infty }{lim}{\displaystyle \sum _{k=1}^n\frac{2n}{n^2+k^2}=\underset{n\to \infty }{lim}\frac{1}{n}}{\displaystyle \sum _{k=1}^n\frac{2}{1+{\left(\frac{k}{n}\right)}^2}}$

$\therefore a={\displaystyle {\int }_0^1\frac{2}{1+x^2}dx=2ta{n}^{-1}x}]{}_0^1=\frac{\pi }{2}$

$f\text{'}\left(\frac{a}{2}\right)=\sqrt[]{2}f\left(\frac{a}{2}\right)$

 $l={\displaystyle \underset{0}{\overset{20\pi }{\int }}{\left(\left|sinx\right|+\left|cosx\right|\right)}^{2}dx=20{\displaystyle \underset{0}{\overset{\pi }{\int }}\left(1+\left|sin2x\right|\right)dx=40{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1+\left|sin2x\right|\right)dx=40{\left(x-\frac{cos2x}{2}\right)}_0^{\frac{\pi }{2}}}}}$

$=40\left(\frac{\pi }{2}+\frac{1}{2}+\frac{1}{2}\right)$ = 20 (p + 2)

 $Letf\left(x\right)=lo{g}_{cosx}cosecx=\frac{logcosecx}{logcosx}$

$f\text{'}\left(x\right)=\frac{logcosx.sinx\left(-cosecxcotx-\left(logcosecx\right)\frac{1}{cosx}.\left(sinx\right)\right)}{{\left(logcosx\right)}^2}$

$Atx=\frac{\pi }{4}f\text{'}\left(\frac{\pi }{4}\right)=\frac{-log\left(\frac{1}{\sqrt[]{2}}\right)+log\sqrt[]{2}}{{\left(log\frac{1}{\sqrt[]{2}}\right)}^2}=\frac{2}{log\sqrt[]{2}}\therefore atx=\frac{\pi }{4},lo{g}_e\left(2f\text{'}\left(x\right)\right)=4$

 $l=60{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(\frac{sin6x-sin4x}{sinx}+\frac{sin4x-sin2x}{sinx}+\frac{sin2x}{sinx}\right)dx}$

$l=60{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(2cos5x+2cos3x+2cosx\right)dx}$

${l=60\left(\frac{2}{5}sin5x+\frac{2}{3}sin3x+2sinx\right)|}_0^{\pi /2}$ = 104

 $l_n\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}\frac{dt}{{\left(t^2+s\right)}^n}}$

Applying integral by parts

$l_n\left(x\right)={\left[\frac{t}{{\left(t^2+5\right)}^n}\right]}_0^x-{\displaystyle \underset{0}{\overset{x}{\int }}n{\left(t^2+5\right)}^{-n-1}.2t^2}$

$10n{l}_{n+1}\left(x\right)+\left(1-2n\right)l_n\left(x\right)=\frac{x}{{\left(x^2+5\right)}^n}$

Put n = 5

$f\left(x\right)={\int }_1^3?\frac{\sqrt[]{x}dx}{(1+x{)}^2}={\int }_1^{\sqrt[]{3}}?\frac{t.2tdt}{{\left(1+t^2\right)}^2}\mathrm{}$ (put $\sqrt[]{x}=t$ )

$={\left(-\frac{\mathrm{t}}{1+{\mathrm{t}}^2}\right)}_1^{\sqrt[]{3}}+{\left({\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?\mathrm{t}\right)}_1^{\sqrt[]{3}}\mathrm{}$ [Applying by parts]

$\begin{array}{rr}& =-\left(\frac{\sqrt[]{3}}{4}-\frac{1}{2}\right)+\frac{\pi }{3}-\frac{\pi }{4}\\ & =\frac{1}{2}-\frac{\sqrt[]{3}}{4}+\frac{\pi }{12}\end{array}$

 $\int {\left(\frac{x}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}\right)}^{2}dx=\int \left(\frac{x}{\mathrm{c}\mathrm{o}\mathrm{s}?x}\right)\cdot \frac{x\mathrm{c}\mathrm{o}\mathrm{s}?xdx}{(x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x{)}^2}$

$\begin{array}{rr}& =\frac{x}{\mathrm{c}\mathrm{o}\mathrm{s}?x}\left(-\frac{1}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}\right)+\int \left(\frac{\mathrm{c}\mathrm{o}\mathrm{s}?x+x\mathrm{s}\mathrm{i}\mathrm{n}?x}{{\mathrm{c}\mathrm{o}\mathrm{s}}^2?x}\right)\left(\frac{1}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}\right)dx\\ & =-\frac{x\mathrm{s}\mathrm{e}\mathrm{c}?x}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}+\int {\mathrm{s}\mathrm{e}\mathrm{c}}^2?xdx\\ & =-\frac{x\mathrm{s}\mathrm{e}\mathrm{c}?x}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}+\mathrm{t}\mathrm{a}\mathrm{n}?x+C\end{array}$