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If \(\displaystyle S = \sum_{n=1}^{\infty}\frac{\sin (n)}{n}.\) Then value of \(\displaystyle 2S+1 = \)

Thanks Papapetros Vaggelis

Thanks Tolaso J. Kos

Calculation of \(\displaystyle \int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx\) **My Try*: Let \(\displaystyle I = \int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx = \int\frac{\sqrt{\tan x}}{\sqrt{\tan x}+1}dx = x+\int\frac{1}{1+\sqrt{\tan x}}dx\) Now Let \(\displaystyle \tan x= t^2\...

Nice Solution Apostolos J. Kos. ** My Try for \(\displaystyle (a)\;\; I = \int\frac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2\cdot (x+1)}dx\) Multiply both \(\bf{N_{r}}\) and \(\bf{D_{r}}\) by \(x+1\), We Get \(\displaystyle I = \int\frac{(x^2-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)^2}dx = \int\frac{\left(1-\...

Evaluation of Following Indefinite Integrals

\(\displaystyle (a)\;\; \int\frac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2\cdot (x+1)}dx\)

\(\displaystyle (b)\;\; \int e^{m\cdot \sin^{-1}(x)}dx\)

Thanking You Grigorios Kostakos , Got it.

**My Try for \(\displaystyle (a)\;\; I = \int_{0}^{\infty}\frac{x^3+3}{x^6\cdot \left(x^2+1\right)}dx\) Now Let \(\displaystyle x = \frac{1}{t}\) and \(\displaystyle dx = -\frac{1}{t^2}\) and Changing Limit, We Get \(\displaystyle I = \int_{0}^{\infty}\frac{(1+3t^3)\cdot t^6\cdot t^2}{t^3\cdot (1+t^...

Thanking You Apostolos J. Kos, Yours (II) Method is very Nice. **My Solution::** Given \(\displaystyle \int_{0}^{\frac{\pi}{2}}\ln (\sin x)dx\) Now Let \(\displaystyle \sin x = \left(\frac{e^{ix}-e^{-ix}}{2i}\right) = \frac{1}{2i}\cdot \left(\frac{e^{2ix}-1}{e^{ix}}\right)\) So Integral Convert into...

Thanking You Grigorios Kostakos Nice Solution.