Math archive (7)

Put $\sin x=t$ . $J_n=2\displaystyle\int_0^1t(1-t)^n\mathrm{d}t\\=2\left(\displaystyle\int_0^1(1-t)^n\mathrm{d}t-\displaystyle\int_0^1\underbrace{(1-t)(1-t)^n}_{(1-t)^{n+1}}\mathrm{d}t\right)$ $= 2\left(\displaystyle\int_0^1t^n\mathrm{d}t-\displaystyle\int_0^1t^{n+1}\mathrm{d}t\right)$ $=2\left(\dfrac{1}{n+1}-\dfrac{1}{n+2}\right)$

$\large\therefore\displaystyle\sum_{n=0}^{\infty}J_n=2\displaystyle\sum_{n=0}^{\infty}\left(\dfrac{1}{n+1}-\dfrac{1}{n+2}\right)$

$\large\mathcal{(A~Telescopic ~Series)}$

$\huge =2(1)=\boxed 2$ $\LARGE\color{#D61F06}{\star\star\star\star\star\star}$

$S = \displaystyle \sum_{n=0}^{\infty} \int_{0}^{\frac{\pi}{2}}\sin(2x)(1-\sin(x))^{n}dx$
$S = \displaystyle \int_{0}^{\frac{\pi}{2}}\sin(2x)\sum_{n=0}^{\infty}(1-\sin(x))^{n}dx$
This is an infinite geometric progression with common ratio $(1-\sin(x))$
$\therefore S = \displaystyle \int_{0}^{\frac{\pi}{2}} \sin(2x)\dfrac{1}{1-(1-\sin(x))}dx = \displaystyle 2\int_{0}^{\frac{\pi}{2}}\cos(x)dx = 2$