calculate-A-0-pi-3-du-1-cos-2-u-3-

Question Number 50421 by Abdo msup. last updated on 16/Dec/18

c a l c u l a t e A = \int_{0}^{\frac{π}{3}} \frac{d u}{{(1 + {c o s}^{2} u)}^{3}}

Commented by maxmathsup by imad last updated on 18/Dec/18

w e h a v e 1 + {c o s}^{2} u = 1 + \frac{1}{1 + {t a n}^{2} u} = \frac{2 + {t a n}^{2} u}{1 + {t a n}^{2} u} \Rightarrow

A = \int_{0}^{\frac{π}{3}} \frac{d u}{{(\frac{2 + {t a n}^{2} u}{1 + {t a n}^{2} u})}^{3}} = \int_{0}^{\frac{π}{3}} \frac{{(1 + {t a n}^{2} u)}^{3}}{{(2 + {t a n}^{2} u)}^{3}} c h a n g e m e n t t a n (u) = x g i v e

A = \int_{0}^{\sqrt{3}} \frac{{(1 + x^{2})}^{3}}{{(2 + x^{2})}^{3}} \frac{d x}{(1 + x^{2})} = \int_{0}^{\sqrt{3}} \frac{x^{4} + 2 x^{2} + 1}{{(x^{2} + 2)}^{3}} d x l e t d e c o m p o s e

F (x) = \frac{x^{4} + 2 x^{2} + 1}{{(x^{2} + 2)}^{3}}

F (x) = G (u) = \frac{u^{2} + 2 u + 1}{{(u + 2)}^{3}} = \frac{a}{u + 2} + \frac{b}{{(u + 2)}^{2}} + \frac{c}{{(u + 2)}^{3}}

c = {l i m}_{u \to - 2} {(u + 2)}^{2} G (u) = 1

{l i m}_{u \to + \infty} u G (u) = 1 = a \Rightarrow G (u) = \frac{1}{u + 2} + \frac{b}{{(u + 2)}^{2}} + \frac{1}{{(u + 2)}^{3}}

G (- 1) = 0 = 1 + b + 1 = b + 2 \Rightarrow b = - 2 \Rightarrow G (u) = \frac{1}{u + 2} - \frac{2}{{(u + 2)}^{2}} + \frac{1}{{(u + 2)}^{3}} \Rightarrow

F (x) = \frac{1}{x^{2} + 2} - \frac{2}{{(x^{2} + 2)}^{2}} + \frac{1}{{(x^{2} + 2)}^{3}} \Rightarrow

\int_{0}^{\sqrt{3}} F (x) d x = \int_{0}^{\sqrt{3}} \frac{d x}{x^{2} + 2} - 2 \int_{0}^{\sqrt{3}} \frac{d x}{{(x^{2} + 2)}^{2}} + \int_{0}^{\sqrt{3}} \frac{d x}{{(x^{2} + 2)}^{3}}

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

\int_{0}^{\sqrt{3}} \frac{d x}{{(x^{2} + 2)}^{2}} =_{x = \sqrt{2} t a n θ} \int_{0}^{a r c t a n (\frac{\sqrt{3}}{\sqrt{2}})} \frac{\sqrt{2} (1 + {t a n}^{2} θ) d θ}{4 {(1 + {t a n}^{2} θ)}^{2}}

= \frac{\sqrt{2}}{4} \int_{0}^{a r c t a n (\frac{\sqrt{3}}{\sqrt{2}})} {c o s}^{2} θ d θ = \frac{\sqrt{2}}{4} \int_{0}^{a r c t a n (\frac{\sqrt{3}}{\sqrt{2}})} \frac{1 + c o s (2 θ)}{2} d θ

= \frac{\sqrt{2}}{8} a r c t a n (\frac{\sqrt{3}}{\sqrt{2}}) + \frac{\sqrt{2}}{16} {[s i n (2 θ)]}_{0}^{a r c t a n (\frac{\sqrt{3}}{\sqrt{2}})}

= \frac{\sqrt{2}}{8} a r c t a n (\frac{\sqrt{3}}{\sqrt{2}}) + \frac{\sqrt{2}}{16} s i n (2 a r c t a n (\frac{\sqrt{3}}{\sqrt{2}})) . \dots b e c o n t i n u e d \dots