find-0-pi-2-dx-3-sinx-

Question Number 40141 by maxmathsup by imad last updated on 16/Jul/18

f i n d \int_{0}^{\frac{π}{2}} \frac{d x}{3 + s i n x}

Commented by maxmathsup by imad last updated on 19/Jul/18

l e t I = \int_{0}^{\frac{π}{2}} \frac{d x}{3 + s i n x} c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e

I = \int_{0}^{1} \frac{1}{3 + \frac{2 t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = 2 \int_{0}^{1} \frac{d t}{3 + 3 t^{2} + 2 t}

= 2 \int_{0}^{1} \frac{d t}{3 t^{2} + 2 t + 3} b u t 3 t^{2} + 2 t + 3 = 3 (t^{2} + \frac{2}{3} t + 1)

= 3 (t^{2} + \frac{2}{3} t + \frac{1}{9} + 1 - \frac{1}{9}) = 3 {{(t + \frac{1}{3})}^{2} + \frac{8}{9}} \Rightarrow

I = \frac{2}{3} \int_{0}^{1} \frac{d t}{{(t + \frac{1}{3})}^{2} + \frac{8}{9}} =_{t + \frac{1}{3} = \frac{2 \sqrt{2}}{3} u} \frac{2}{3} \int_{\frac{1}{2 \sqrt{2}}}^{\sqrt{2}} \frac{1}{\frac{8}{9} (1 + u^{2})} \frac{2 \sqrt{2}}{3} d u

= \frac{4 \sqrt{2}}{9} \frac{9}{8} \int_{\frac{1}{2 \sqrt{2}}}^{\sqrt{2}} \frac{d u}{1 + u^{2}} = \frac{\sqrt{2}}{2} {[a r c t a n u]}_{\frac{1}{\sqrt{2}}}^{\sqrt{2}} \Rightarrow

I = \frac{\sqrt{2}}{2} {a r c t a n (\sqrt{2}) - a r c t a n (\frac{1}{\sqrt{2}})} .