Question

integral 0 to π (x sin x/1+sinx)dX
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Answer 1

Step-by-step explanation:

We have,

$i = \int_{0}^{\pi} \frac{x \sin(x) }{1 + \sin(x) } dx \\ ........(i)$

$\implies \: i = \int_{0}^{\pi} \frac{(\pi - x) \sin(\pi - x) }{1 + \sin(\pi - x) } dx$

$\implies \: i = \int_{0}^{\pi} \frac{(\pi - x) \sin(x) }{1 + \sin(x) } dx \\ ....(ii)$

Adding (i) and (ii),

$\implies \: 2i = \int_{0}^{\pi} \frac{\pi \sin(x) }{1 + \sin(x) } dx$

$\implies \: 2i = \pi \int_{0}^{\pi} \frac{1 + \sin(x) - 1 }{1 + \sin(x) } dx$

$\implies \frac{2}{\pi} i = \int_{0}^{\pi}dx - \int_{0}^{\pi} \frac{dx}{1 + \sin(x) }$

$\implies \frac{2}{\pi} i = \pi - \int_{0}^{\pi} \frac{ \sec^{2} ( \frac{x}{2} ) }{(1 + \tan( \frac{x}{2} ))^{2} } dx$

$\implies \frac{2}{\pi} i = \pi -2 \int _{0}^{\pi} \frac{ \frac{1}{2} \sec^{2} ( \frac{x}{2} ) dx }{(1 + \tan( \frac{x}{2} ))^{2} }$

$\implies \frac{2}{\pi} i = \pi -2 \int _{1}^{ \infty } \frac{dt}{ {t}^{2} }$

$\implies \frac{2}{\pi} i = \pi + 2 [ \frac{1}{t} ] _{1}^{ \infty }$

$\implies \frac{2}{\pi} i = \pi -2(1 - \frac{1}{ \infty } )$

$\implies \frac{2}{\pi} i = \pi -2$

$\implies i = \frac{ (\pi -2)\pi }{ 2 }$

$\implies \: i = \frac{ {\pi}^{2} }{2} - \pi$