Integrals

${\displaystyle \int \frac{sinx}{si{n}^{3}x+co{s}^{3}x}dx}$

$I={\displaystyle \int \frac{tanxse{c}^{2}x}{ta{n}^{3}x+1}dx}$

Let, tan x = t

sec² xdx = dt

${\displaystyle \int \frac{tdt}{t^3+1}={\displaystyle \int \frac{t}{\left(t+1\right)\left(t^2-t+1\right)}dt}}$    

$\frac{t}{\left(t+1\right)\left(t^2-t+1\right)}=\frac{A}{\left(t+1\right)}+\frac{Bt+c}{t^2-t+1}$

$l=\frac{-1}{3}{\displaystyle \int \frac{1}{t+1}dt+\frac{1}{3}{\displaystyle \int \frac{t+1}{t^2-t+1}dt}}$

The value of 18 $\left(\alpha +\beta +{\gamma }^2\right)$ is equal to 3.

$S_n={\displaystyle \sum _{r=1}^{n-1}r\left(n-r\right)}$

$=n\sum r-\sum r^2$

$=n\frac{\left(n-1\right)n}{2}-\frac{\left(n-1\right)n\left(2n-1\right)}{6}=\frac{n^3-n}{6}$

$S={\displaystyle \sum _{n=4}^{\infty }\left(\frac{2}{6}.\frac{n^3-n}{n!}-\frac{1}{\left(n-2\right)!}\right)=\frac{1}{3}}{\displaystyle \sum _{n=4}^{\infty }\frac{1}{\left(n-3\right)!}=\frac{e-1}{3}}$

$l={\pi }^2{\displaystyle \underset{0}{\overset{2}{\int }}\left(sin\left(\frac{\pi x}{2}\right){\left(x-\left[x\right]\right)}^{\left[x\right]}\right)dx}$

$l={\pi }^2{\displaystyle \underset{0}{\overset{1}{\int }}sin\left(\frac{\pi x}{2}\right).x^{0}dx+{\pi }^2}{\displaystyle \underset{1}{\overset{2}{\int }}sin\left(\frac{\pi x}{2}\right){\left(x-1\right)}^{1}dx}$

$l={\pi }^2\left(\frac{2}{\pi }\right)+\frac{2{\pi }^2}{\pi }\left[1-0\right]+2\pi \times \frac{2}{\pi }{\left[sin\frac{\pi x}{2}\right]}_1^2$

$\begin{array}{l}l=2\pi +2\pi +4\left(0-1\right)\\ l=4\pi -4\end{array}$

$\frac{S_{10}}{S_P}=\frac{100}{P^2}$

$S_P=\frac{S_{10}.P^2}{100}\Rightarrow S_{11}=\frac{S_{10}.121}{100}$

$\frac{a_{11}}{a_{10}}=\frac{S_{11}-S_{10}}{S_{10}-S_9}=\frac{S_{10}.\frac{121}{100}-S_{10}}{S_{10}-\frac{S_{10}.81}{100}}$

= $\frac{21}{19}$

$\underset{x\to 0}{lim}\frac{si{n}^2\left(\pi co{s}^{4}x\right)}{x^4}=\underset{x\to 0}{lim}\frac{si{n}^2\left(\pi -\pi co{s}^{4}x\right)}{x^4}=\underset{x\to 0}{lim}\frac{si{n}^2\left(\pi si{n}^{2}x\left(1+co{s}^{2}x\right)\right)}{x^4}$

$=\underset{x\to 0}{lim}\frac{si{n}^2\left(\pi si{n}^{2}x\left(1+co{s}^{2}x\right)\right)}{{\pi }^{2}si{n}^{4}x{\left(1+co{s}^{2}x\right)}^2}\times \frac{{\pi }^{2}si{n}^{4}x{\left(1+co{s}^{2}x\right)}^2}{x^4}=4{\pi }^2\underset{x\to 0}{lim}\frac{si{n}^{4}x}{x^4}=4{\pi }^2$

Let $t=\frac{3}{2}{x}^2+\frac{5}{3}{x}^3+\frac{7}{4}{x}^4+......$

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

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

$=2\frac{x^2}{1-x}+ln\left(1-x\right)+x=\frac{x^2+x}{1-x}+ln\left(1-x\right)=\frac{x\left(1+x\right)}{1-x}+ln\left(1-x\right)$              

$=2\frac{x^2}{1-x}+ln\left(1-x\right)+x=\frac{x^2+x}{1-x}+ln\left(1-x\right)=\frac{x\left(1+x\right)}{1-x}+ln\left(1-x\right)$

$y^{1/4}+\frac{1}{y^{1/4}}=2x$

${\left(y^{1/4}\right)}^2-2x{y}^{1/4}+1=0$    

=> ${\left(y^{1/4}\right)}^2-2x{y}^{1/4}+1=0$

$\Rightarrow \frac{dy}{dx}=\frac{4y}{\sqrt[]{x^2-1}}..........\left(i\right)$

$\Rightarrow \left(x^2-1\right)\frac{d^{2}y}{d{x}^2}+\frac{xdy}{dx}-16y=0from\left(i\right)$      

=>α = 1, β = -16

|α - β| = 17

$l={\displaystyle \underset{0}{\overset{1}{\int }}\frac{\sqrt[]{x}dx}{\left(1+x\right)\left(1+3x\right)\left(3+x\right)}}$

Put x = t² then dx = 2tdt

$l={\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{\left(1+t^2\right)\left(3+t^2\right)}-{\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{\left(1+3t^2\right)\left(3+t^2\right)}}}$

$=\frac{1}{2}{\left(ta{n}^{-1}t\right)}_0^1-\frac{3}{8\sqrt[]{3}}{\left(ta{n}^{-1}\frac{t}{\sqrt[]{3}}\right)}_0^1-\frac{8}{8\sqrt[]{3}}{\left(ta{n}^{-1}\sqrt[]{3}t\right)}_0^1$

$=\frac{\pi }{8}\left(1-\frac{\sqrt[]{3}}{2}\right)$      

$n\left(n+1\right)x^2+2\left(2n+1\right)x+4$

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

${\displaystyle \sum _{n=1}^9\frac{x}{\left(nx+2\right)\left(\left(n+1\right)x+2\right)}}$

$={\displaystyle \sum _{n=1}^9\left[\frac{1}{nx+2}-\frac{1}{\left(n+1\right)x+2}\right]}$

$\left[\left(\frac{1}{x+2}-\frac{1}{2x+2}\right)+\left(\frac{1}{2x+2}-\frac{1}{3x+2}\right)+....+\left(\frac{1}{9x+2}-\frac{1}{10x+2}\right)\right]$

$\therefore \underset{x\to 2}{Lt}\frac{9x}{\left(10x+2\right)\left(x+2\right)}=\frac{9}{44}$              

${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left[\frac{1+si{n}^{2}x}{1+{\pi }^{sinx}}+\frac{1+si{n}^{2}x}{1+{\pi }^{-sinx}}\right]dx}$

$={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1+si{n}^{2}x\right)dx=\frac{\pi }{2}+\frac{1}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1-cos2x\right)dx}}$

$=\frac{\pi }{2}+\frac{1}{2}{\left[x-\frac{sin2x}{2}\right]}_0^{\frac{\pi }{2}}$
$=\frac{\pi }{2}+\frac{\pi }{4}=\frac{3\pi }{4}$