let-f-0-pi-4-dx-1-sin-sinx-with-0-lt-lt-pi-2-1-explicite-f-2-calculate-0-pi-4-dx-1-sin-sinx-2-

Question Number 78273 by msup trace by abdo last updated on 15/Jan/20

l e t f (θ) = \int_{0}^{\frac{π}{4}} \frac{d x}{1 + s i n θ s i n x}

w i t h 0 < θ < \frac{π}{2}

1) e x p l i c i t e f (θ)

2) c a l c u l a t e \int_{0}^{\frac{π}{4}} \frac{d x}{{(1 + s i n θ s i n x)}^{2}}

Commented by mathmax by abdo last updated on 18/Jan/20

1) c h a n g e m e n t t a n (\frac{x}{2}) = u \Rightarrow f (θ) = \int_{0}^{\sqrt{2} - 1} \frac{1}{1 + s i n θ \times \frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}

= \int_{0}^{\sqrt{2} - 1} \frac{2 d u}{1 + u^{2} + 2 s i n θ u} = \int_{0}^{\sqrt{2} - 1} \frac{2 d u}{u^{2} + 2 s i n θ u + {s i n}^{2} θ + {c o s}^{2} θ}

= \int_{0}^{\sqrt{2} - 1} \frac{2 d u}{{(u + s i n θ)}^{2} + {c o s}^{2} θ} =_{u + s i n θ = (c o s θ) t} 2 \int_{t a n θ}^{\frac{\sqrt{2} - 1}{c o s θ} + t a n θ} \frac{c o s θ d t}{{c o s}^{2} θ (1 + t^{2})}

= \frac{2}{c o s θ} {[a r c t a n t]}_{t a n θ}^{\frac{\sqrt{2} - 1}{c o s θ} + t a n θ} \Rightarrow f (θ) = \frac{2}{c o s θ} {a r c t a n (\frac{\sqrt{2} - 1}{c o s θ} + t a n θ) - θ}