calculate-A-0-pi-2-dx-2-cos-sinx-pi-lt-lt-pi-

Question Number 78271 by msup trace by abdo last updated on 15/Jan/20

c a l c u l a t e A_{θ} = \int_{0}^{\frac{π}{2}} \frac{d x}{2 + c o s θ s i n x}

- π < θ < π

Commented by mathmax by abdo last updated on 19/Jan/20

A (θ) = \int_{0}^{\frac{π}{2}} \frac{d x}{2 + c o s θ s i n x} \Rightarrow A (θ) =_{t a n (\frac{x}{2}) = t} \int_{0}^{1} \frac{2 d t}{(1 + t^{2}) (2 + c o s θ \frac{2 t}{1 + t^{2}})}

= \int_{0}^{1} \frac{2 d t}{2 + 2 t^{2} + 2 t c o s θ} = \int_{0}^{1} \frac{d t}{t^{2} + t c o s θ + 1} = \int_{0}^{1} \frac{d t}{t^{2} + 2 t \frac{c o s θ}{2} + \frac{{c o s}^{2} θ}{4} + 1 - \frac{{c o s}^{2} θ}{4}}

= \int_{0}^{1} \frac{d t}{{(t + \frac{c o s θ}{2})}^{2} + \frac{4 - {c o s}^{2} θ}{4}} =_{t + \frac{c o s θ}{2} = \frac{\sqrt{4 - {c o s}^{2} θ}}{2} u \Rightarrow u = \frac{2 t + c o s θ}{\sqrt{4 - {c o s}^{2} θ}}}

= \int_{\frac{c o s θ}{\sqrt{4 - {c o s}^{2} θ}}}^{\frac{2 + c o s θ}{\sqrt{4 - {c o s}^{2} θ}}} \frac{1}{\frac{4 - {c o s}^{2} θ}{4} (1 + u^{2})} \times \frac{\sqrt{4 - {c o s}^{2} θ}}{2} d u

= 2 \int_{\frac{c o s θ}{\sqrt{4 - {c o s}^{2} θ}}}^{\frac{2 + c o s θ}{\sqrt{4 - {c o s}^{2} θ}}} \frac{d u}{1 + u^{2}} = 2 {[a r c t a n (u)]}_{\frac{c o s θ}{\sqrt{4 - {c o s}^{2} θ}}}^{\frac{2 + c o s θ}{\sqrt{4 - {c o s}^{2} θ}}}

= 2 {a r c t a n (\frac{2 + c o s θ}{\sqrt{4 - {c o s}^{2} θ}}) - a r c t a n (\frac{c o s θ}{\sqrt{4 - {c o s}^{2} θ}})} .