Maths Integrals

 $x^4-2x^3+2x-1={\left(x-1\right)}^2\left(x^2-1\right)$

$sin\pi x=sin(\pi \left(1-x\right)$

$=-sin\left(sin\pi \left(x-1\right)\right)$

$\underset{x\to 1}{lim}\frac{\left(x^2-1\right)\cdot si{n}^2\pi x}{\left(x^2-1\right){\left(x-1\right)}^2}=\underset{x\to 1}{lim}\frac{si{n}^2\left(\pi \left(x-1\right)\right)}{{\left(x-1\right)}^2}$

$=\underset{x\to 1}{lim}\frac{si{n}^2\left(\pi \left(x-1\right)\right)}{{\left(\pi \left(x-1\right)\right)}^2}\cdot {\pi }^2$

= 2

$?{\mathcal{l}}_{1}and{\mathcal{l}}_2$ are perpendicular, so

$3*1+\left(-2\right)\left(\frac{\alpha }{2}\right)+0*2=0$

a = 3

Now angle between ${\mathcal{l}}_{2}and{\mathcal{l}}_3$ ,

$cos\theta =\frac{1\left(-3\right)+\frac{\alpha }{2}\left(-2\right)+2\left(4\right)}{\sqrt[]{1+\frac{{\alpha }^2}{4}+4}.\sqrt[]{9+4+16}}$

$\Rightarrow cos\theta =\frac{\frac{2}{29}}{2}\Rightarrow \theta =co{s}^{-1}\left(\frac{4}{29}\right)=se{c}^{-1}\left(\frac{29}{4}\right)$

$\underset{x\to \frac{1}{\sqrt[]{2}}}{lim}\frac{sin\left(co{s}^{-1}x\right)-x}{1-tan\left(co{s}^{-1}x\right)}$

let $co{s}^{-1}x=\frac{\pi }{4}+\theta$

= $\underset{\theta \to \infty }{lim}\frac{\text{exception 25:}sin\theta }{-2tan\theta }\left(1-tan\theta \right)=-$$\frac{1}{}$

$cot\left({\displaystyle {\sum }_{n=1}^{50}ta{n}^{-1}}\left(\frac{1}{1+n+n^2}\right)\right)$

$=cot\left(ta{n}^{-1}51-ta{n}^{-1}1\right)$

$=cot\left(co{t}^{-1}\left(\frac{52}{50}\right)\right)$

$=\frac{26}{25}$

${\displaystyle {\int }_0^1\frac{1}{7^{\left(\frac{1}{x}\right)}}dx,let\frac{1}{x}=t}$

$\frac{-1}{x^2}dx=dt$

$={\displaystyle {\int }_{\infty }^1\frac{1}{-t^2{7}^{\left|t\right|}}dt={\displaystyle {\int }_1^{\infty }\frac{1}{t^2{7}^{\left|t\right|}}dt}}$

$=\frac{1}{7}{\left[-\frac{1}{t}\right]}_1^2+\frac{1}{7^2}{\left[\frac{-1}{t}\right]}_2^3+\frac{1}{7^3}{\left[-\frac{1}{t}\right]}_2^3+...$

$={\displaystyle {\sum }_{n=1}^{\infty }\frac{1}{7_n}\left(\frac{1}{n}-\frac{1}{n+1}\right)}$

$=1+6log\frac{6}{7}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{1+cosx}{x+sinx}}dx\\ Putx+sinx=t\Rightarrow \left(1+cosx\right)dx=dt\\ \therefore I={\displaystyle \int \frac{dt}{t}}=log\left|t\right|=log\left|x+sinx\right|+C\\ Hence,therequiredsolutionislog\left|x+sinx\right|+C.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{e^{6logx}-e^{5logx}}{e^{4logx}-e^{3logx}}}dx\\ \therefore I={\displaystyle \int \frac{e^{log{x}^6}-e^{log{x}^5}}{e^{log{x}^4}-e^{log{x}^3}}}dx\\ ={\displaystyle \int \frac{x^6-x^5}{x^4-x^3}dx}={\displaystyle \int \frac{x^2\left(x^4-x^3\right)}{x^4-x^3}dx}={\displaystyle \int x^{2}dx}\\ =\frac{1}{3}{x}^3+C\\ Hence,therequiredsolutionis\frac{1}{3}{x}^3+C.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{\left(x^2+2\right)}{x+1}}dx\\ \therefore I={\displaystyle \int \left[\left(x-1\right)+\frac{3}{x+1}\right]}dx\\ ={\displaystyle \int \left(x-1\right)dx}+3{\displaystyle \int \frac{1}{x+1}dx}\\ =\frac{x^2}{2}-x+3log\left|x+1\right|+C\\ Hence,therequiredsolutionis\frac{x^2}{2}-x+3log\left|x+1\right|+C.\end{array}$