let F(x) =∫_0 ^x ((1+sint)/(2+cost))dt 1) find a explicite form of f(x) 2) calculate ∫


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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
$1)changement tan((t/2))=u give F(x)=∫_0 ^(tan((x/2))) ((1+((2u)/(1+u^2 )))/(2+((1−u^2 )/(1+u^2 )))) ((2du)/(1+u^2 )) =∫_0 ^(tan((x/2))) ((1+u^2 +2u)/(2+2u^2 +1−u^2 )) ((2du)/(1+u^2 )) =2 ∫_0 ^(tan((x/2))) ((u^2 +2u +1)/((u^2 +1)(u^2 +3))) du let decompose F(u) = ((u^2 +2u +1)/((u^2 +1)(u^2 +3))) ⇒F(u) =((au +b)/(u^2 +1)) +((cu +d)/(u^2 +3)) ⇒(u^2 +3)(au+b) +(u^2 +1)(cu +d)=u^2 +2u +1 ⇒ au^3 +bu^2 +3au +3b +cu^3 +du^2 +cu +d =u^2 +2u +1 ⇒ (a+c)u^3 +(b+d)u^2 +(3a+c)u +3b +d =u^2 +2u +1⇒ a+c=0,b+d =1,3a+c =2 ,3b +d =1 ⇒ c=−a ⇒3a−a =2 ⇒a =1 wehave d=1−b ⇒ 3b +1−b =1 ⇒b=0 ⇒d=1 ⇒ F(u)=(u/(u^2 +1)) +((−u +1)/(u^2 +3)) ⇒∫ F(u)du =(1/2)ln(u^2 +1) −(1/2)ln(u^2 +3) +arctan(u) +c ⇒ F(x)=2 [(1/2)ln(u^2 +1)−(1/2)ln(u^2 +3) +arctan(u)]_0 ^(tan((x/2))) =2 {(1/2)ln(1+tan^2 ((x/2)))−(1/2)ln(3+tan^2 ((x/2)))+(x/2)} F(x)=x +ln(1+tan^2 ((x/2)))−ln(3+tan^2 ((x/2)))$
$1) c h a n g e m e n t t a n (\frac{t}{2}) = u g i v e$ $F (x) = \int_{0}^{t a n (\frac{x}{2})} \frac{1 + \frac{2 u}{1 + u^{2}}}{2 + \frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}$ $= \int_{0}^{t a n (\frac{x}{2})} \frac{1 + u^{2} + 2 u}{2 + 2 u^{2} + 1 - u^{2}} \frac{2 d u}{1 + u^{2}}$ $= 2 \int_{0}^{t a n (\frac{x}{2})} \frac{u^{2} + 2 u + 1}{(u^{2} + 1) (u^{2} + 3)} d u l e t d e c o m p o s e$ $F (u) = \frac{u^{2} + 2 u + 1}{(u^{2} + 1) (u^{2} + 3)} \Rightarrow F (u) = \frac{a u + b}{u^{2} + 1} + \frac{c u + d}{u^{2} + 3}$ $\Rightarrow (u^{2} + 3) (a u + b) + (u^{2} + 1) (c u + d) = u^{2} + 2 u + 1 \Rightarrow$ ${a u}^{3} + {b u}^{2} + 3 a u + 3 b + {c u}^{3} + {d u}^{2} + c u + d = u^{2} + 2 u + 1 \Rightarrow$ $(a + c) u^{3} + (b + d) u^{2} + (3 a + c) u + 3 b + d = u^{2} + 2 u + 1 \Rightarrow$ $a + c = 0, b + d = 1, 3 a + c = 2, 3 b + d = 1 \Rightarrow$ $c = - a \Rightarrow 3 a - a = 2 \Rightarrow a = 1 w e h a v e d = 1 - b \Rightarrow$ $3 b + 1 - b = 1 \Rightarrow b = 0 \Rightarrow d = 1 \Rightarrow$ $F (u) = \frac{u}{u^{2} + 1} + \frac{- u + 1}{u^{2} + 3} \Rightarrow \int F (u) d u$ $= \frac{1}{2} l n (u^{2} + 1) - \frac{1}{2} l n (u^{2} + 3) + a r c t a n (u) + c \Rightarrow$ $F (x) = 2 {[\frac{1}{2} l n (u^{2} + 1) - \frac{1}{2} l n (u^{2} + 3) + a r c t a n (u)]}_{0}^{t a n (\frac{x}{2})}$ $= 2 {\frac{1}{2} l n (1 + {t a n}^{2} (\frac{x}{2})) - \frac{1}{2} l n (3 + {t a n}^{2} (\frac{x}{2})) + \frac{x}{2}}$ $F (x) = x + l n (1 + {t a n}^{2} (\frac{x}{2})) - l n (3 + {t a n}^{2} (\frac{x}{2}))$
Commented by Abdo msup. last updated on 05/Apr/19

$2) w e h a v e \int_{0}^{π} \frac{1 + s i n t}{2 + c o s t} d t =_{t a n (\frac{x}{2})} \int_{0}^{\infty} \frac{1 + \frac{2 u}{1 + u^{2}}}{2 + \frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}$ $= 2 \int_{0}^{\infty} \frac{u^{2} + 2 u + 1}{(u^{2} + 1) (u^{2} + 3)} d u$ $= {[l n (u^{2} + 1) - l n (u^{2} + 3) + 2 a r c t a n (u)]}_{0}^{+ \infty}$ $= {[l n (\frac{u^{2} + 1}{u^{2} + 3}) + 2 a r c t a n (u)]}_{0}^{+ \infty} = π - l n (\frac{1}{3})$ $= π + l n (3) .$
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