0-pi-sin-x-dx-1-sin-x-

Question Number 90997 by jagoll last updated on 27/Apr/20

\int \underset{0}{\overset{π}{}} \frac{\sin x d x}{1 + \sin x}

Commented by john santu last updated on 27/Apr/20

Commented by jagoll last updated on 27/Apr/20

t h a n k y o u

Commented by mathmax by abdo last updated on 27/Apr/20

I = \int_{0}^{π} \frac{s i n x}{1 + s i n x} d x \Rightarrow I = \int_{0}^{π} \frac{1 + s i n x - 1}{1 + s i n x} d x

= π - \int_{0}^{π} \frac{d x}{1 + s i n x} w e h a v e \int_{0}^{π} \frac{d x}{1 + s i n x} =_{t a n (\frac{x}{2}) = t} \int_{0}^{\infty} \frac{2 d t}{(1 + t^{2}) (1 + \frac{2 t}{1 + t^{2}})}

{= \int_{0}^{\infty} \frac{2 d t}{1 + t^{2} + 2 t} = \int_{0}^{\infty} \frac{2 d t}{{(t + 1)}^{2}} = - \frac{2}{t + 1}]}_{0}^{\infty} = 2 \Rightarrow I = π - 2