Integrals

Given limit
$={\displaystyle \underset{0}{\overset{2}{\int }}\frac{dx}{1+4x^2}}$

$={\left[\frac{1}{2}ta{n}^{-1}2x\right]}_0^2=\frac{1}{2}ta{n}^{-1}4$

$l={\displaystyle \underset{-\frac{1}{\sqrt[]{2}}}{\overset{\frac{1}{\sqrt[]{2}}}{\int }}{\left(\frac{16x^2}{{\left(x^2-1\right)}^2}\right)}^{\frac{1}{2}}dx}$

$=2{\displaystyle \underset{0}{\overset{\frac{1}{\sqrt[]{2}}}{\int }}\frac{4x}{1-x^2}dx}$

$=4ln2$

Required area

$={\displaystyle \underset{-2}{\overset{4}{\int }}\left(\frac{3x}{2}+6-\frac{3x^2}{4}\right)dx=27}$

$\frac{3x^2}{4}=\frac{3x}{2}+6\Rightarrow x=-2,4$

$2sin\frac{\pi }{8}sin\frac{2\pi }{8}tan\frac{3\pi }{8}sin\left(\pi -\frac{5\pi }{8}\right)sin\left(\pi -\frac{6\pi }{8}\right)sin\left(\pi -\frac{7\pi }{8}\right)$

$={\left(\frac{1}{\sqrt[]{2}}\right)}^2.2si{n}^2\frac{\pi }{8}si{n}^2\frac{3\pi }{8}$

^{$=\frac{1}{4}.{\left(2sin\frac{\pi }{8}sin\frac{3\pi }{8}\right)}^2$}

$=\frac{1}{4}.{\left(2sin\frac{\pi }{8}sin\frac{3\pi }{8}\right)}^2$

${\displaystyle \underset{0}{\overset{5}{\int }}\frac{x+\left[x\right]}{e^{x-\left[x\right]}}}dx$

$={\displaystyle \sum _{r=1}^5{\displaystyle \underset{r-1}{\overset{r}{\int }}\frac{\left(x+1-r\right)}{e^{x+1-r}}}dx}$

Put x + 1 – r = t ->dx = dt

$=5{\left[-t{e}^{-t}-e^{-t}\right]}_0^1$              

$=5\left[-1e^{-1}-e^{-1}+e^0\right]=5\left(1-\frac{2}{e}\right)$

$\Rightarrow \alpha =-10\beta =5\Rightarrow {\left(\alpha +\beta \right)}^2=25$

I(x) = ${\displaystyle \int \frac{se{c}^{2}x-2022}{si{n}^{2022}x}dx}$

$={\displaystyle \int si{n}^{-2022}x\cdot se{c}^{2}xdx-{\displaystyle \int 2022si{n}^{-2022}xdx}}$

$=si{n}^{-2022}x\cdot tanx-{\displaystyle \int \left(-2022\right)si{n}^{-2023}x\cdot cosx\cdot tanxdx-{\displaystyle \int 2022si{n}^{-2022}xdx}}$

$=tanx\cdot si{n}^{-2022}x+2022{\displaystyle \int si{n}^{-2022}-2022{\displaystyle \int si{n}^{-2022}xdx}}$

I(x) = $tanx\cdot si{n}^{-2022}x+c$

Given, $I\left(\frac{\pi }{4}\right)=2^{1011}$

$2^{1011}=\frac{1}{{\left(\frac{1}{\sqrt[]{2}}\right)}^{2022}}+c\Rightarrow c=0$

$I\left(x\right)=\frac{tanx}{si{n}^{2022}x},I\left(\frac{\pi }{3}\right)=\frac{\sqrt[]{3}}{\left(\frac{\sqrt[]{3}}{2}\right)}=\frac{2^{2022}}{{\left(\sqrt[]{3}\right)}^{2021}}$

  ${\displaystyle \underset{3}{\overset{b}{\int }}\frac{1}{3}\left(\frac{1}{x^2-4}+\frac{1}{x^2-1}\right)}dx$

$=\frac{1}{3}{\left[\frac{1}{4}\mathcal{l}n\frac{x-2}{x+2}-\frac{1}{2}\mathcal{l}n\frac{x-1}{x+1}\right]}_3^6$

$=\mathcal{l}n{\left(\frac{49}{40}\right)}^{\frac{1}{12}}$

$\frac{5\left(b-2\right)}{b+2}/\frac{4{\left(b-1\right)}^2}{{\left(b+1\right)}^2}=\frac{49}{40}$

$\frac{4b}{2\left(b^2+1\right)}=\frac{-b+198}{99b-2}$

->b³ – 3b – 198 = 0

b = 6

  $ta{n}^{-1}\left(\frac{\frac{1}{\sqrt[]{2}-1}}{\frac{1}{\sqrt[]{2}}}\right)$

$=ta{n}^{-1}\left(1-\sqrt[]{2}\right)$

$=ta{n}^{-1}\left(\sqrt[]{2}-1\right)$

$=-\theta$

$tan\theta =\sqrt[]{2}-1$

$tan2\theta =\frac{2\left(\sqrt[]{2}-1\right)}{1-{\left(\sqrt[]{2}-1\right)}^2}$

$=\frac{2\left(\sqrt[]{2}-1\right)}{2\left(\sqrt[]{2}-1\right)}$

tan 2 θ = 1

$\theta =\frac{\pi }{8}$                

$2sin12°-sin72°$

$=sin12°-\left(sin72°-sin12°\right)$

$=sin12°-2cos42°sin30°$

$=2cos30°\left(-sin18°\right)$

$=-\sqrt[]{3}sin18°$

$=-\sqrt[]{3}\left(\frac{\sqrt[]{5}-1}{4}\right)$  

$\underset{x\to \frac{\pi }{2}}{lim}\frac{si{n}^{2}x\left(si{n}^{2}x-3sinx+2\right)}{co{s}^{2}x.\left(\sqrt[]{2si{n}^{2}x+3sinx+4}+\sqrt[]{si{n}^{2}x+6sinx+2}\right)}$  

$\downarrow$                                                       $\downarrow$                            

3                                                       3

$\underset{x\to \frac{\pi }{2}}{lim}\frac{1}{9}.\frac{\left(sinx-1\right)\left(sinx-2\right)}{\left(1-sinx\right)\left(1+sinx\right)}=\frac{1}{6}\left(-1\right).\frac{-1}{2}=\frac{1}{12}$