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Maximum value of $$\int_{0}^{y}\sqrt{x^4+\left(y-y^2\right)^2}dx\;,$$ Where \(y\in \left[0,1\right]\)

My Try for (1):: Let \(\displaystyle I = \int_{0}^{\pi}\lfloor \cot x \rfloor dx\;,\) Now using \(\displaystyle \int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dx\) So we get \(\displaystyle I = \int_{0}^{\pi}\lfloor -\cot x \rfloor dx\;,\) So we get \(\displaystyle I = \int_{0}^{\pi}\left(\lfloor \cot x \rf...

(1) Evaluation of \(\displaystyle \int_{0}^{\pi}\lfloor \cot x \rfloor dx\;,\) where \(\lfloor x \rfloor \) is floor function of \(x\).

(2) Evaluation of \(\displaystyle \int_{0}^{\pi}\lfloor \cos x \rfloor dx\;,\) where \(\lfloor x \rfloor \) is floor function of \(x\).

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

$(b)\;$ Evaluation of $\displaystyle \int_{0}^{1}\frac{\ln(1+x^2)}{1+x}dx$

$(a)\;\;$ Evaluation of $\displaystyle \int \sqrt{\tan^2 x-3}\; dx$

$(b)\;\;$ Evaluation of $\displaystyle \int \sqrt{\tan x+3}\; dx$

Evaluation of \(\displaystyle \int e^{x\sin x+\cos x}\cdot \left(\frac{x^4\cos^3 x-x\sin x+\cos x}{x^2\cos^2 x}\right)dx\)

Let $$S = \sum _{ m=1 }^{ \infty }{ \sum _{ n=1 }^{ \infty }{ \frac { { m }^{ 2 }n }{ { 3 }^{ m }(n{ 3 }^{ m }+m{ 3 }^{ n }) } } } $$ Using the fact that $$\sum_{m = 1}^\infty\sum_{n = 1}^\infty f(m,n) = \sum_{m = 1}^\infty\sum_{n = 1}^\infty f(n,m) $$ we can rewrite \(S\) as $$S = \sum _{ m=1 }^{ \...

If \(\displaystyle y'(x)+y(x)\cdot g'(x) = g(x)\cdot g'(x)\;,y(0)=0\;,x\in \mathbb{R}\;,\) If \(\displaystyle f'(x) = \frac{d}{dx}\left(f(x)\right)\) and \(g(x)\) is a non constant differentiable function on \(\displaystyle \mathbb{R}\) with \(g(0) = g(2) = 0\;\). Then the value of \(y(2)\) is?

Grigorios Kostakos wrote:Jacks, your result for \((2)\,(a)\) it is not correct: Assuming \(a\neq b\) you must have that: \[\displaystyle \int\frac{\sin^2 x}{a^2\sin^2 x+b^2\cos^2 x}dx=\frac{1}{a^2-b^2}\Bigl({x-\frac{b}{a}\arctan\bigl({\tfrac{a}{b}\tan{x}}\bigr)}\Bigr)\]

Thanking You Grigorios Kostakos, Got my mistake.

I have evaluate \((a)::\) Given \(\displaystyle \int\frac{\sin^2 x}{a^2\sin^2 x+b^2\cos^2 x}dx = \frac{1}{a^2}\int \frac{a^2\sin^2 x}{a^2\sin^2x+b^2\cos^2 x}dx\) Divide both \(\bf{N_{r}}\) and \(\bf{D_{r}}\) by \(a^2\cos^2 x\;,\) We Get \(\displaystyle =\frac{1}{a^2}\int\frac{\tan^2 x}{\tan^2 x+A}dx...