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Evaluation of \(\displaystyle \int_{25\pi/4}^{53\pi/4} \frac{1}{(1+2^{\sin x})(1+2^{\cos x})}\,dx\)

Evaluation of \(\displaystyle \int\frac{\cot^2 x}{1+\tan^3 x}dx\)

Let \(\displaystyle S_{n} = \sum_{k=1}^{n}\frac{n}{n^2+nk+k^2}\) and \(\displaystyle T_{n} = \sum_{k=0}^{n-1}\frac{n}{n^2+nk+k^2}\) for \(n=1,2,3,4.....\) Then which one is Right \(\displaystyle (a):: S_{n}<\frac{\pi}{3\sqrt{3}}\) \(\displaystyle (b):: S_{n}>\frac{\pi}{3\sqrt{3}}\) \(\displaystyle (...

(1) \(\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\sin^9 x}{\sin^3 x+\cos^3 x}dx\) (2) \(\displaystyle \int_{0}^{a}x^3\cdot \sqrt{2ax-x^2}dx\) (3) \(\displaystyle \int_{0}^{1}\frac{2-x^2}{(1+x)\sqrt{1-x^2}}dx\) (4) \(\displaystyle\int_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}\left(\frac{x^4}{1-x^4}...

Evaluation of \(\displaystyle \int_{0}^{1}\frac{x^2\ln(x)}{\sqrt{1-x^2}}dx\)

Papapetros Vaggelis wrote:Question : What \(\displaystyle{\mathrm{IIT\,\, JEE\,\, 2014}}\) is ; Where \(\displaystyle{\mathrm{IIT\,\, JEE\,\, 2014}}\) takes place ;

To Papapetros Vaggelis, Actually It is a paper for Engg. entrance exam for B.Tech(from IIT,NIT) in INDIA.
It Contain 3 papers, Physics,Chemistry and Maths.

The value of \(\displaystyle \int_{0}^{1}4x^3\cdot \left\{\frac{d^2}{dx^2}\left(1-x^2\right)^5\right\}dx = \)



Asked in \(\bf{IIT\; JEE\; 2014}\)

If \(f(0) = 2\) and \(f^{'}(0) = 3\) and \(f^{''}(x) = f(x)\). Then value of \(f(4) =\)

If \(\displaystyle I_{n} = \int\frac{x^n}{(ax^2+bx+c)^{\frac{1}{2}}}dx\) and \(\displaystyle n\in \mathbb{N}\). Then value of \(\displaystyle I_{n+1}\) in terms of \(\displaystyle I_{n}\) and \(\displaystyle I_{n-1}\)

If \(\displaystyle y^2 = ax^2+2bx+c\;,\) and \(\displaystyle U_{n} = \int \frac{x^n}{y}dx\;,\) Then prove that \((n+1)\,a\,U_{n}+(2n+1)\,b\,U_{n}+c\,U_{n-1}=x^n\,y\)

and deduce that \(a\,U_{1} = y-b\,U_{0}\) and \(\displaystyle 2a^2U_{2} = y\,(ax-3b)-(ac-3b^2)\,U_{0}\)