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$\lim_{n\to\infty}\frac{1}{n^2}+\frac{2}{n^2}+.........+\frac{n}{n^2}$

= $\lim_{n\to\infty}\frac{1}{n}\bigg(\frac{1}{n}+\frac{2}{n}+.........+\frac{n}{n}\bigg)$

= $\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^{n}\frac{r}{n}$

= $\int_0^1 x.dx$

= $\frac{1}{2}$

is the answer tanx=1 ???

go to this link :::

http://www.goiit.com/posts/list/algebra-solve-for-definite-salutes-32483.htm

$\mbox{REDUCTION FORMULAES:::::}$

$1.\;\;I_n=\int sin^nx.dx$

$I_n=\frac{-(sinx)^{n-1}.cosx}{n}+\frac{(n-1)}{n}.I_{n-2}$

$2.\;\;I_{m,n}=\int x^m(1-x)^n.dx$

$I_{m,n}=\frac{n}{m+1}.I_{m+1,n+1}$

specially , $\int_0^1 x^m(1-x)^n.dx=\frac{m!\;n!}{(m+n+1)!}$

$3.\;\;I_n=\int cos^nx.dx$

$I_n=\frac{(cosx)^{n-1}.sinx}{n}+\frac{(n-1)}{n}I_{n-2}$

$4.\;\;I_n=\int\frac{dx}{(x^2+a^2)^n}$

$2(n-1)a^2I_n=\frac{x}{(x^2+a^2)^{n-1}}+(2n-3)I_{n-1}$

$5.\;\;I_{m,n}=\int x^m.cosnx.dx$

$I_{m,n}=\frac{x^m.sin nx}{n}+\frac{m.x^{m-1}.cosnx}{n^2}-\frac{m(m-1)}{n^2}$

$6.\;\;I_{m,n}=\int cos^m x.cos nx.dx$

$(m+n)I_{m,n}=cos^m x.sin nx+m.I_{m-1,n-1}$

$7.\;\;I_m=\int_0^{2a} x^m\sqrt{2ax-x^2}.dx$

$I_m=\frac{(2m+1).a}{m+2}.I_{m-1}$

$8.\;\;I_n=\int cot^n x.dx$

$I_n+I_{n-2}=\frac{(cotx)^{n-1}}{n-1}$

$9.\;\;I_n=\int\frac{x^n.dx}{\sqrt{ax^2+2bx+c}}$

$(n+1).a.I_{n+1}+(2n+1).b.I_n+n.c.I_{n-1}=x^n\sqrt{ax^2+2bx+c}$

$10.\;\;I_n=\int \frac{dx}{x^n\sqrt{ax+b}}$

$I_n=\frac{\sqrt{ax+b}}{(n-1).b.x^{n-1}}-\bigg(\frac{2n-3}{2n-1}\bigg).\bigg(\frac{a}{b}\bigg).I_{n-1}$

$11.\;\;I_n=\int e^{\alpha x}.sin^n x.dx$

$I_n=\frac{e^{\alpha x}}{\alpha^2+n^2}.(sinx)^{n-1}(\alpha sinx-n cosx)+\frac{n(n-1)}{\alpha^2+n^2}.I_{n-2}$

$12.\;\;\int_0^{\infty}\frac{dx}{(x+\sqrt{x^2+1})^n}=\frac{n}{n^2-1}$

$13.\;\;\int_a^b\sqrt{(x-a)(b-x)}.dx=\frac{\pi}{8}.(a-b)^2$

$14.\;\;\int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}}=\pi$

$15.\;\;\int_a^b\sqrt{\frac{x-a}{b-x}}.dx=\frac{\pi}{2}.(b-a)$

P.S : these formulaes were not copied from any source..... i had compiled these formulaes from various books...... hope it is useful for you.......

property 4 :

$\int_a^bf(x).dx=\int_a^bf(a+b-x).dx$

let $I=\int_a^b f(x).dx$

let a+b-x=t

dx=-dt

$I=-\int_b^a f(a+b-t).dt$

$I=\int_a^b f(a+b-t).dt$

$I=\int_a^b f(a+b-x).dx$ (since variable is immaterial in definite integration)

so $\int_a^bf(x).dx=\int_a^bf(a+b-x).dx$

chennai super kings!!!!!!!!!!!

$I_1=\int\limits_{sin^2t}^{1+cos^2t}x.f(x(2-x)).dx$

apply property 4

$I_1=\int\limits_{sin^2t}^{1+cos^2t}(2-x).f(x(2-x)).dx$

$I_1=\int\limits_{sin^2t}^{1+cos^2t}2f(x(2-x)).dx-\int\limits_{sin^2t}^{1+cos^2t}xf(x(2-x)).dx$

I_1=2I_2-I_1

$\frac{I_1}{I_2}=1$

$\int_a^bf(x)dx=\int_0^1(b-a).f((b-a)x+a).dx$

$\int_{-4}^{-5}e^{(x+5)^2}dx=\int_0^1(-5+4)e^{(x-1)^2}.dx$

$\int_{-4}^{-5}e^{(x+5)^2}dx=-\int_0^1e^{(x-1)^2}.dx$ .............(1)

$\int_{\frac{1}{3}}^{\frac{2}{3}}e^{9(x-\frac{2}{3})^2}dx=\int_0^1\frac{1}{3}.e^{9(\frac{x+1}{3}-\frac{2}{3})^2}.dx$

$3.\int_{\frac{1}{3}}^{\frac{2}{3}}e^{9(x-\frac{2}{3})^2}dx=\int_0^1e^{(x-1)^2}.dx$ ..........(2)

adding (1) and (2)

$\int_{-4}^{-5}e^{(x+5)^2}dx+3\int_{\frac{1}{3}}^{\frac{2}{3}}e^{9(x-\frac{2}{3})^2}.dx=0$

hence proved

it is used to estimate the hardness of water.....

hard water contains Ca2+ and Mg2+ ions ..... these ions form complex with EDTA.....

we use a borate buffer of pH=9.2 , eriochrome black T as indicator......

$cos^{-1}\frac{x}{a}+cos^{-1}\frac{y}{b}=\theta$

$cos^{-1}\Bigg(\frac{xy}{ab}-\sqrt{1-\frac{x^2}{a^2}}.\sqrt{1-\frac{y^2}{b^2}}\Bigg)=\theta$

$\frac{xy}{ab}-\sqrt{1-\frac{x^2}{a^2}}.\sqrt{1-\frac{y^2}{b^2}}=cos\theta$

$\frac{xy}{ab}-cos\theta=\sqrt{1-\frac{x^2}{a^2}}.\sqrt{1-\frac{y^2}{b^2}}$

square both sides

$\Bigg(\frac{xy}{ab}-cos\theta\Bigg)^2=\Bigg(1-\frac{x^2}{a^2}\Bigg)\Bigg(1-\frac{y^2}{b^2}\Bigg)$

$\frac{x^2y^2}{a^2b^2}-\frac{2xycos\theta}{ab}+cos^2\theta=1-\frac{x^2}{a^2}-\frac{y^2}{b^2}+\frac{x^2y^2}{a^2b^2}$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2xycos\theta}{ab}=1-cos^2\theta$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2xycos\theta}{ab}=sin^2\theta$

$sinA=\frac{a}{2R}$

$cosA=\frac{b^2+c^2-a^2}{2bc}$

$cotA=\frac{cosA}{sinA}=\frac{2R(b^2+c^2-a^2)}{2abc}$

$cotA=\frac{b^2+c^2-a^2}{4\bigtriangleup}$

similarly

$cotB=\frac{a^2+c^2-b^2}{4\bigtriangleup}$

$cotC=\frac{a^2+b^2-c^2}{4\bigtriangleup}$

$\frac{cotC}{cotA+cotB}=\frac{a^2+b^2-c^2}{b^2+c^2-a^2+a^2+c^2-b^2}$

$\frac{cotC}{cotA+cotB}=\frac{a^2+b^2-c^2}{2c^2}$

$\frac{cotC}{cotA+cotB}=\frac{2007c^2-c^2}{2c^2}$

$\frac{cotC}{cotA+cotB}=1003$

PS: please type the question properly without any errors..... you have typed cotA/(cotB+cotA)

let y = $\frac{\alpha^3}{2}cosec^2\bigg(\frac{1}{2}tan^{-1}\frac{\alpha}{\beta}\bigg)+\frac{\beta^3}{2}sec^2\bigg(\frac{1}{2}tan^{-1}\frac{\alpha}{\beta}\bigg)$

let $\frac{1}{2}tan^{-1}\frac{\alpha}{\beta}=\theta$

y = $\frac{\alpha^3}{2}cosec^2\theta+\frac{\beta^3}{2}sec^2\theta$

y = $\frac{\alpha^3}{2sin^2\theta}+\frac{\beta^3}{2cos^2\theta}$

y = $\frac{\alpha^3}{1-cos2\theta}+\frac{\beta^3}{1+cos2\theta}$

y = $\frac{\alpha^3}{1-cos\bigg(tan^{-1}\frac{\alpha}{\beta}\bigg)}+\frac{\beta^3}{1+cos\bigg(tan^{-1}\frac{\alpha}{\beta}\bigg)}$

y = $\frac{\alpha^3}{1-cos\bigg(cos^{-1}\frac{\beta}{\sqrt{\alpha^2+\beta^2}}\bigg)}+\frac{\beta^3}{1+cos\bigg(cos^{-1}\frac{\beta}{\sqrt{\alpha^2+\beta^2}}\bigg)}$

y = $\frac{\alpha^3}{1-\frac{\beta}{\sqrt{\alpha^2+\beta^2}}}+\frac{\beta^3}{1+\frac{\beta}{\sqrt{\alpha^2+\beta^2}}}$

y = $\frac{\alpha^3\sqrt{\alpha^2+\beta^2}}{\sqrt{\alpha^2+\beta^2}-\beta}+\frac{\beta^3\sqrt{\alpha^2+\beta^2}}{\sqrt{\alpha^2+\beta^2}+\beta}$

y = $\sqrt{\alpha^2+\beta^2}.\Bigg(\frac{\alpha^3(\sqrt{\alpha^2+\beta^2}+\beta)}{\alpha^2+\beta^2-\beta^2}+\frac{\beta^3(\sqrt{\alpha^2+\beta^2}-\alpha)}{\alpha^2+\beta^2-\alpha^2}\Bigg)$

y = $\sqrt{\alpha^2+\beta^2}.\bigg(\alpha(\sqrt{\alpha^2+\beta^2}+\beta)+\beta(\sqrt{\alpha^2+\beta^2}-\alpha)\bigg)$

y = $\alpha(\alpha^2+\beta^2)+\beta(\alpha^2+\beta^2)$

y = $(\alpha+\beta)(\alpha^2+\beta^2)$

let y = $tan^{-1}\Bigg(\frac{xcos\theta}{1-xsin\theta}\Bigg)-cot^{-1}\Bigg(\frac{cos\theta}{x-sin\theta}\Bigg)$

y = $tan^{-1}\Bigg(\frac{xcos\theta}{1-xsin\theta}\Bigg)-tan^{-1}\Bigg(\frac{x-sin\theta}{cos\theta}\Bigg)$

y = $tan^{-1}\Bigg(\frac{\frac{xcos\theta}{1-xsin\theta}-\frac{x-sin\theta}{cos\theta}}{1+\frac{x(x-sin\theta)}{1-xsin\theta}}\Bigg)$

y = $tan^{-1}\Bigg(\frac{xcos^2\theta-x+sin\theta+x^2sin\theta-xsin^2\theta}{cos\theta(1-xsin\theta+x^2-xsin\theta)}\Bigg)$

y = $tan^{-1}\Bigg(\frac{x(1-sin^2\theta)-x+sin\theta+x^2sin\theta-xsin^2\theta}{cos\theta(1-xsin\theta+x^2-xsin\theta)}\Bigg)$

y = $tan^{-1}\Bigg(\frac{-xsin^2\theta+sin\theta+x^2sin\theta-xsin^2\theta}{cos\theta(1-xsin\theta+x^2-xsin\theta)}\Bigg)$

y = $tan^{-1}\Bigg(\frac{sin\theta(1+x^2-2xsin\theta)}{cos\theta(1+x^2-2xsin\theta)}\Bigg)$

y = $tan^{-1}(tan\theta)$

y = $\theta$

yes we can have limits from 2 to 3.......
but the method remains the same.....

nuclear density = nuclear mass/volume of nucleus

nuclear density = $\frac{A.m}{\frac{4}{3}\pi R^3}$

where A is mass number

m - average mass of a nucleon

R - radius of nucleus

now we know the relation $R=R_0 A^{\frac{1}{3}}$ substituting in the nuclear density

nuclear density = $\frac{A.m}{\frac{4}{3}.\pi.R_0^3.A}$

nuclear density = $\frac{m}{\frac{4}{3}.\pi.R_0^3}$ = constant

so nuclear density is same for all elements.......