Question

$\:\:\color{red} \displaystyle \rm \int \limits_{ \sum \limits_{n = 0}^{ \infty } \frac{( - 1 {)}^{n} }{(2n + 1)!} {\pi}^{2n + 1} }^{ \sum \limits_{n = 1}^{ \infty } \frac{1}{n} } \left( \lim_{n \to \infty } \bigg (1 - \frac{1}{n} \bigg)^{n} \right)^{ \frac{d}{dx} \left( \frac{ {x}^{2} }{ \sin^{2} (x) + \cos^{2} (x) } \right) } \: dx$​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\sf{\int\limits^{\sum\limits^{\infty}_{n=1}\frac{1}{n}}_{\sum\limits^{\infty}_{n=0}\frac{\left(-1\right)^{n}}{(2n+1)!}\cdot\,{\pi}^{2n+1}}\left\{\lim\limits_{n\to\infty}\left(1-\dfrac{1}{n}\right)^{n}\right\}^{\dfrac{d}{dx}\left(\dfrac{x^2}{sin^2(x)+cos^2(x)}\right)}\,dx}$

$\displaystyle=\sf{\int\limits^{\sum\limits^{\infty}_{n=1}\frac{1}{n}}_{\sum\limits^{\infty}_{n=0}\frac{\left(-1\right)^{n}}{(2n+1)!}\cdot\,{\pi}^{2n+1}}\left\{\lim\limits_{n\to\infty}\left(1+\dfrac{-1}{n}\right)^{n}\right\}^{\dfrac{d}{dx}\left(x^2\right)}\,dx}$

$\displaystyle=\sf{\int\limits^{\sum\limits^{\infty}_{n=1}\frac{1}{n}}_{\sum\limits^{\infty}_{n=0}\frac{\left(-1\right)^{n}}{(2n+1)!}\cdot\,{\pi}^{2n+1}}\left[\lim\limits_{n\to\infty}\left\{\left(1+\dfrac{1}{-n}\right)^{-n}\right\}^{-1}\right]^{2x}\,dx}$

$\displaystyle=\sf{\int\limits^{\sum\limits^{\infty}_{n=1}\frac{1}{n}}_{\sum\limits^{\infty}_{n=0}\frac{\left(-1\right)^{n}}{(2n+1)!}\cdot\,{\pi}^{2n+1}}\left\{\lim\limits_{n\to\infty}\left(1+\dfrac{1}{-n}\right)^{-n}\right\}^{-2x}\,dx}$

We know,  $\,\,\boxed{\displaystyle\blue{\tt{\lim_{x\to\infty}\left(1+\dfrac{1}{x}\right)^{x}=e}}}$

So,

$\displaystyle=\sf{\int\limits^{\sum\limits^{\infty}_{n=1}\frac{1}{n}}_{\sum\limits^{\infty}_{n=0}\frac{\left(-1\right)^{n}}{(2n+1)!}\cdot\,{\pi}^{2n+1}}\left\{e\right\}^{-2x}\,dx}$

Now, let us consider the upper and lower limits of the integral

UPPER LIMIT:

$\displaystyle\rm{\sum^{\infty}_{n\to\infty}\dfrac{1}{n}=\infty}$

LOWER LIMIT:

$\displaystyle\sum^{\infty}_{n=0}\dfrac{\left(-1\right)^n}{(2n+1)!}\cdot\,\pi^{2n+1}$

$\displaystyle=\dfrac{\left(-1\right)^{0}}{1!}\cdot\,\pi^{1}+\dfrac{\left(-1\right)^{1}}{3!}\cdot\pi^{3}+\dfrac{\left(-1\right)^{2}}{5!}\cdot\pi^{5}+\dfrac{\left(-1\right)^{3}}{7!}\cdot\pi^{7}+...$

$\displaystyle=\pi-\dfrac{\pi^{3}}{3!}+\dfrac{\pi^{5}}{5!}-\dfrac{\pi^{7}}{7!}+...$

$=\sin(\pi)$

$=0$

So,

$\displaystyle=\sf{\int\limits^{\infty}_{0}\left\{e\right\}^{-2x}\,dx}$

$\displaystyle=\sf{\left[\dfrac{e^{-2x}}{-2}\right]^{\infty}_{0}}$

$\displaystyle=\sf{\dfrac{e^{-\infty}}{-2}-\dfrac{e^{0}}{-2}}$

$\displaystyle=\sf{0-\dfrac{1}{-2}}$

$\displaystyle=\sf{\dfrac{1}{2}}$