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Thanks seanieg89 for Nice explanation.

$Calculation of $\displaystyle \frac{\tan 20^0+\tan 40^0+\tan 80^0-\tan 60^0}{\sin 40^0}$ $What i have try::, Using If $A+B+C= \pi\;,$ Then $\tan(A+B) = \tan(\pi-C)$ $So we get $\tan A+\tan B+\tan C = \tan A\cdot \tan B\cdot \tan C$ $And Using $\tan (60^0-A)\cdot \tan(A)\cdot...

Re: MX2 2016 Integration Marathon $Let $I = \int\frac{1}{(x+2)^{\frac{6}{5}}\cdot (x-7)^{\frac{4}{5}}} = \int\frac{1}{\left(\frac{x-7}{x+2}\right)^{\frac{4}{5}}\cdot (x+2)^2}dx$ $Now Put $\left(\frac{x-7}{x+2}\right) = t\;,$ $Then $\displaystyle \left[\frac{(x+2)-9}{x+2}\right] =...

$All positive integer ordered pairs $(x,y)$ for which $\displaystyle \binom{x}{y} = 2016$$

Re: MX2 2016 Integration Marathon $let $I = \int\frac{1}{\left(x+\sqrt{x^2-1}\right)^{2}}dx$\;, Now put $(x+\sqrt{x^2-1}) = t\;,$ $Then $(x+\sqrt{x^2-1}) dx = \sqrt{x^2-1}dt$ $Now Using $(x+\sqrt{x^2-1})\cdot (x-\sqrt{x^2-1})=1 \;,$ Then $(x-\sqrt{x^2-1}) =\frac{1}{t}$ $So we get...

$An ellipse with length of major axis $4$ and length of minor axis $2$$ $touches the coordinate axis , Then how can we prove that locus of $ $its center is circle $x^2+y^2 = 5$$

$Prove that the number of rational points on the Circumference of a circle$ $having center $(\pi,e)$ is atmost $1$. $ Given a point $a$ and $b$ is rational ,if $a$ and $b$ both are rational number $

Re: MX2 2015 Integration Marathon $(1)\;\; Evaluation of $\int_{0}^{\pi}\frac{x}{\sqrt{1+\sin^3 x}}((3\pi \cos x+4\sin x)\sin^2 x+4)dx = $ $(2)\;\; If $L = \lim_{n\rightarrow \infty}\frac{1}{n}\sum^{n}_{k=1}k\ln\left[\frac{n^2+(k-1)^2}{n^2+k^2}\right]$ exists, Then $L=$ $(3)\;\; If $F(n)...

Re: MX2 2015 Integration Marathon $Let $I(a) = \int_{0}^{\pi}\frac{1}{a-\cos \theta } d\theta\;,$ Now Put $\cos \theta = \frac{1-\tan^2 \frac{\theta}{2}}{1+\tan^2 \frac{\theta}{2}}\;,$ we get $ $So $I(a) = \int_{0}^{\pi}\frac{\sec^2 \frac{\theta}{2}}{(a-1)+(a+1)\tan^2...

Re: MX2 2015 Integration Marathon $(a)\;\;Evaluation of $\displaystyle \bf{\int_{\frac{\pi}{2}}^{\frac{5\pi}{2}}\frac{e^{\tan^{-1}(\sin x)}}{e^{\tan^{-1}(\sin x)}+e^{\tan^{-1}(\cos x)}}dx}$ $(b)\;Determine positive integer $n\leq 5,$ such that $\int_{0}^{1}e^x\cdot (x-1)^ndx = 16-6e$

Re: MX2 2015 Integration Marathon $Let, $\text{V(n)}=\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin^2 x} \text{d}x$ $and $\text{J}= \text{V(n) - V(n-1)}=\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\sin^2nx-\sin^2(n-1)x}{\sin^2 x} \text{d}x$ =\displaystyle...

Re: MX2 2015 Integration Marathon $Let $I = \int\left(\frac{\tan^{-1} x}{x-\tan^{-1} x}\right)^2 dx =\int\left(\frac{\tan^{-1} x}{\tan^{-1} x-x}\right)^2 dx $ $So we can write $I = \int\left(1+\frac{x}{\tan^{-1} x-x}\right)^2dx$ $So we get $I =\int 1dx+\int\frac{2x}{\tan^{-1}...

Re: MX2 2015 Integration Marathon Sorry Manager I have edited my post.

Re: MX2 2015 Integration Marathon $(1)\; If $f(x) = x^3+3x+4\;,$ Then value of $\int_{-1}^{1}f(x)dx+\int_{0}^{4}f^{-1}(x)dx = $ $(2)\; Let $I_{n} = \int_{0}^{\frac{\pi}{2}}x(\cos x+\sin x)^ndx\;,$ Then value of $\frac{101I_{101}-\frac{\pi}{2}}{I_{99}} = $ $(3)\; Let $f(k) =...

Re: MX2 2015 Integration Marathon $Let $I = \int_{0}^{1}\frac{\ln(1+x)}{1+x^2}dx\;,$ Now put $x=\frac{1-t}{1+t} = -\left(\frac{t-1}{t+1}\right) = \frac{2}{1+t}-1$ $So we get $dx = -\frac{2}{(1+t)^2}dt\;,$ and changing limit\;, we get$ I =...

Re: MX2 2015 Integration Marathon $Let $\displaystyle I = \int\frac{x^2}{(x\sin x+\cos x)^2}dx$ $We can write $\displaystyle (x\sin x+\cos x) = \sqrt{1+x^2}\left\{\frac{x}{\sqrt{1+x^2}}\cdot \sin x+\frac{1}{\sqrt{1+x^2}}\cdot \cos x\right\}$ $ = \sqrt{1+x^2}\cdot \cos \left(x-\phi...

Re: MX2 2015 Integration Marathon (a)\;\; \int_{0}^{1}\binom{207}{7}x^{200}\cdot (1-x)^{7}dx\;\;\;\;\; (b)\;\; \int_{0}^{\ln 2}\frac{2e^{3x}+e^{2x}-1}{e^{3x}+e^{2x}-e^x+1}dx $(c)\;\; If $V_{n} = \int_{0}^{\frac{\pi}{2}}\frac{\sin^2(nx) }{\sin^2 x}dx\;,$ Where $n\in \mathbb{N}\;,$ Then...

Re: MX2 2015 Integration Marathon $Let $I = \int\frac{a^2\sin^2 x+b^2\cos^2 x}{a^4\sin^2 x+b^4\cos^2 x}dx$\;, Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\cos^2 x.$ $So\; we\; get $I = \int\frac{a^2 \tan^2 x+b^2}{a^4\tan^2 x+b^4}dx$ $Now\; Put \; $\tan x=t\;,$ Then $\sec^2 dx =...

Re: MX2 2015 Integration Marathon $Let $I = \int\frac{1}{\sin^ 6+\cos^6 x}dx = \int\frac{1}{1-3\sin^2 x\cdot \cos^2 x}dx = \int\frac{4}{4-3\sin^2 (2x)}dx$ $Now\; Divide\; both\; Numerator\; and \; Denominator\; by\; $\cos^2(2x),$ We\; get$ $$I = \int\frac{4\sec^2 2x}{4(1+\tan^2...

Re: MX2 2015 Integration Marathon $Evaluation of $\displaystyle \int\frac{2x^3-1}{x^6+2x^3+9x^2+1}dx\;\; $ and $\;\; \displaystyle \int\frac{x^2\cos^2 x-x\sin x-\cos x-1}{(1+x\sin x)^2}dx$$