0-pi-2-1-2-cos-x-dx-

Question Number 53295 by gunawan last updated on 20/Jan/19

\int_{0}^{\frac{π}{2}} \frac{1}{2 + \cos x} d x = \dots

Commented by maxmathsup by imad last updated on 20/Jan/19

c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e

A = \int_{0}^{\frac{π}{2}} \frac{d x}{2 + c o s x} = \int_{0}^{1} \frac{1}{2 + \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = \int_{0}^{1} \frac{2 d t}{2 + 2 t^{2} + 1 - t^{2}} = 2 \int_{0}^{1} \frac{d t}{3 + t^{2}}

=_{t = \sqrt{3} u} 2 \int_{0}^{\frac{1}{\sqrt{3}}} \frac{\sqrt{3} d u}{3 (1 + u^{2})} = \frac{2 \sqrt{3}}{3} \int_{0}^{\frac{1}{\sqrt{3}}} \frac{d u}{1 + u^{2}} = \frac{2 \sqrt{3}}{3} a r c t a n (\frac{1}{\sqrt{3}}) = \frac{2}{\sqrt{3}} \frac{π}{6}

= \frac{π}{3 \sqrt{3}} \Rightarrow ★ I = \frac{π}{3 \sqrt{3}} ★

Answered by tanmay.chaudhury50@gmail.com last updated on 20/Jan/19

\int_{0}^{\frac{π}{2}} \frac{1}{2 + \frac{1 - {t a n}^{2} \frac{x}{2}}{1 + {t a n}^{2} \frac{x}{2}}} d x

\int_{0}^{\frac{π}{2}} \frac{{s e c}^{2} \frac{x}{2}}{2 + 2 {t a n}^{2} \frac{x}{2} + 1 - {t a n}^{2} \frac{x}{2}} d x

t = t a n \frac{x}{2} d t = \frac{1}{2} {s e c}^{2} \frac{x}{2} d x

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

2 \times \frac{1}{\sqrt{3}} ∣ {t a n}^{- 1} (\frac{t}{\sqrt{3}}) ∣_{0}^{1}

\frac{2}{\sqrt{3}} [{t a n}^{- 1} (\frac{1}{\sqrt{3}})] = \frac{2}{\sqrt{3}} \times \frac{π}{6} = \frac{π}{3 \sqrt{3}}

Commented by gunawan last updated on 20/Jan/19

thank you Sir

Facebook Tweet Pin