let-F-x-0-x-1-sint-2-cost-dt-1-find-a-explicite-form-of-f-x-2-calculate-0-pi-1-sint-2-cost-dt-

Question Number 57417 by Abdo msup. last updated on 03/Apr/19

l e t F (x) = \int_{0}^{x} \frac{1 + s i n t}{2 + c o s t} d t

1) f i n d a e x p l i c i t e f o r m o f f (x)

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

Commented by Abdo msup. last updated on 05/Apr/19

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

F (x) = \int_{0}^{t a n (\frac{x}{2})} \frac{1 + \frac{2 u}{1 + u^{2}}}{2 + \frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}

= \int_{0}^{t a n (\frac{x}{2})} \frac{1 + u^{2} + 2 u}{2 + 2 u^{2} + 1 - u^{2}} \frac{2 d u}{1 + u^{2}}

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

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

\Rightarrow (u^{2} + 3) (a u + b) + (u^{2} + 1) (c u + d) = u^{2} + 2 u + 1 \Rightarrow

{a u}^{3} + {b u}^{2} + 3 a u + 3 b + {c u}^{3} + {d u}^{2} + c u + d = u^{2} + 2 u + 1 \Rightarrow

(a + c) u^{3} + (b + d) u^{2} + (3 a + c) u + 3 b + d = u^{2} + 2 u + 1 \Rightarrow

a + c = 0, b + d = 1, 3 a + c = 2, 3 b + d = 1 \Rightarrow

c = - a \Rightarrow 3 a - a = 2 \Rightarrow a = 1 w e h a v e d = 1 - b \Rightarrow

3 b + 1 - b = 1 \Rightarrow b = 0 \Rightarrow d = 1 \Rightarrow

F (u) = \frac{u}{u^{2} + 1} + \frac{- u + 1}{u^{2} + 3} \Rightarrow \int F (u) d u

= \frac{1}{2} l n (u^{2} + 1) - \frac{1}{2} l n (u^{2} + 3) + a r c t a n (u) + c \Rightarrow

F (x) = 2 {[\frac{1}{2} l n (u^{2} + 1) - \frac{1}{2} l n (u^{2} + 3) + a r c t a n (u)]}_{0}^{t a n (\frac{x}{2})}

= 2 {\frac{1}{2} l n (1 + {t a n}^{2} (\frac{x}{2})) - \frac{1}{2} l n (3 + {t a n}^{2} (\frac{x}{2})) + \frac{x}{2}}

F (x) = x + l n (1 + {t a n}^{2} (\frac{x}{2})) - l n (3 + {t a n}^{2} (\frac{x}{2}))

Commented by Abdo msup. last updated on 05/Apr/19

2) w e h a v e \int_{0}^{π} \frac{1 + s i n t}{2 + c o s t} d t =_{t a n (\frac{x}{2})} \int_{0}^{\infty} \frac{1 + \frac{2 u}{1 + u^{2}}}{2 + \frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}

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

= {[l n (u^{2} + 1) - l n (u^{2} + 3) + 2 a r c t a n (u)]}_{0}^{+ \infty}

= {[l n (\frac{u^{2} + 1}{u^{2} + 3}) + 2 a r c t a n (u)]}_{0}^{+ \infty} = π - l n (\frac{1}{3})

= π + l n (3) .