let f(x) =∫ (dt/(x +cost +cos(2t))) (x real) 1) find a explicit form of f(x) 2)determine also ∫ (dt/((x+cost +cos(2t))^2 )) 3) find ∫ (dt/(1+cos(t)+cos(2t))) and ∫ (dt/((3 +cos(t)+cos(2t))^2 ))


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Question Number 58488 by Mr X pcx last updated on 23/Apr/19

$l e t f (x) = \int \frac{d t}{x + c o s t + c o s (2 t)} (x r e a l)$ $1) f i n d a e x p l i c i t f o r m o f f (x)$ $2) d e t e r m i n e a l s o \int \frac{d t}{{(x + c o s t + c o s (2 t))}^{2}}$ $3) f i n d \int \frac{d t}{1 + c o s (t) + c o s (2 t)} a n d$ $\int \frac{d t}{{(3 + c o s (t) + c o s (2 t))}^{2}}$
Commented by maxmathsup by imad last updated on 25/Apr/19
$1) we have f(x) =∫ (dt/(x+cost +2cos^2 t−1)) let put cost =u and decompose F(u) =(1/(2u^2 +u +x−1)) ⇒Δ =1−8(x−1) =9−8x case 1 9−8x<0 ⇒no roots and (1/(2u^2 +u +x−1)) =(1/(2{u^2 +(u/2) +((x−1)/2)})) =(1/(2{ u^2 +2(1/4)u +(1/(16)) +((x−1)/2)−(1/(16))})) =(1/(2{ (u+(1/2))^2 +((8x−9)/(16))})) ⇒f(x) =∫ (dt/(2{(cost +(1/2))^2 +((8x−9)/(16))})) changement cost +(1/2) =((√(8x−9))/4)u give 4cost +2 =(√(8x−9))u ⇒ u =(1/(√(8x−9)))(4cost +2) ⇒du =((−4 sint)/(√(8x−9)))dt =((−4(√(1−cos^2 t)))/(√(8x−9))) dt ⇒ (dt/du) =−((√(8x−9))/(4(√(1−(((√(8x−9))/4)u−(1/2))^2 )))) ⇒ f(x) = ∫ ((√(8x−9))/(8(√(1−(((√(8x−9))/4)u−(1/2))^2 ))(((8x−9)/(16)))(1+u^2 ))) du ...be continued... case 2 if 9−8x>0 ⇒F(u) have two poles u_1 =((−1+(√(9−8x)))/4) u_2 =((−1−(√(9−8x)))/4) and F(u) =(1/(2(u−u_1 )(u−u_2 ))) =(1/(2(u_1 −u_2 ))){(1/(u−u_1 )) −(1/(u−u_2 ))} = (1/(2(((√(9−8x))/2)))){ (1/(u−u_1 )) −(1/(u−u_2 ))} ⇒ f(x) =(1/(√(9−8x))){ ∫ (dt/(cost −u_1 )) −∫ (dt/(cost −u_2 ))}let find ∫ (dt/(cost −a)) ch. tan((t/2)) =u give ∫ (dt/(cost −a)) =∫ (1/(((1−u^2 )/(1+u^2 )) −a)) ((2du)/(1+u^2 )) = ∫ ((2du)/(1−u^2 −a(1+u^2 ))) =∫ ((2du)/(1−u^2 −a−au^2 )) =∫ ((2du)/(1−a−(1+a)u^2 )) =∫ ((2du)/((1+a)u^2 +a−1)) ...$
$1) w e h a v e f (x) = \int \frac{d t}{x + c o s t + 2 {c o s}^{2} t - 1} l e t p u t c o s t = u a n d d e c o m p o s e$ $F (u) = \frac{1}{2 u^{2} + u + x - 1} \Rightarrow Δ = 1 - 8 (x - 1) = 9 - 8 x c a s e 1 9 - 8 x < 0 \Rightarrow n o r o o t s$ $a n d \frac{1}{2 u^{2} + u + x - 1} = \frac{1}{2 {u^{2} + \frac{u}{2} + \frac{x - 1}{2}}} = \frac{1}{2 {u^{2} + 2 \frac{1}{4} u + \frac{1}{16} + \frac{x - 1}{2} - \frac{1}{16}}}$ $= \frac{1}{2 {{(u + \frac{1}{2})}^{2} + \frac{8 x - 9}{16}}} \Rightarrow f (x) = \int \frac{d t}{2 {{(c o s t + \frac{1}{2})}^{2} + \frac{8 x - 9}{16}}}$ $c h a n g e m e n t c o s t + \frac{1}{2} = \frac{\sqrt{8 x - 9}}{4} u g i v e 4 c o s t + 2 = \sqrt{8 x - 9} u \Rightarrow$ $u = \frac{1}{\sqrt{8 x - 9}} (4 c o s t + 2) \Rightarrow d u = \frac{- 4 s i n t}{\sqrt{8 x - 9}} d t = \frac{- 4 \sqrt{1 - {c o s}^{2} t}}{\sqrt{8 x - 9}} d t \Rightarrow$ $\frac{d t}{d u} = - \frac{\sqrt{8 x - 9}}{4 \sqrt{1 - {(\frac{\sqrt{8 x - 9}}{4} u - \frac{1}{2})}^{2}}} \Rightarrow$ $f (x) = \int \frac{\sqrt{8 x - 9}}{8 \sqrt{1 - {(\frac{\sqrt{8 x - 9}}{4} u - \frac{1}{2})}^{2}} (\frac{8 x - 9}{16}) (1 + u^{2})} d u . . . b e c o n t i n u e d . . .$ $c a s e 2 i f 9 - 8 x > 0 \Rightarrow F (u) h a v e t w o p o l e s u_{1} = \frac{- 1 + \sqrt{9 - 8 x}}{4}$ $u_{2} = \frac{- 1 - \sqrt{9 - 8 x}}{4} a n d F (u) = \frac{1}{2 (u - u_{1}) (u - u_{2})}$ $= \frac{1}{2 (u_{1} - u_{2})} {\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}}} = \frac{1}{2 (\frac{\sqrt{9 - 8 x}}{2})} {\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}}} \Rightarrow$ $f (x) = \frac{1}{\sqrt{9 - 8 x}} {\int \frac{d t}{c o s t - u_{1}} - \int \frac{d t}{c o s t - u_{2}}} l e t f i n d \int \frac{d t}{c o s t - a}$ $c h . t a n (\frac{t}{2}) = u g i v e \int \frac{d t}{c o s t - a} = \int \frac{1}{\frac{1 - u^{2}}{1 + u^{2}} - a} \frac{2 d u}{1 + u^{2}}$ $= \int \frac{2 d u}{1 - u^{2} - a (1 + u^{2})} = \int \frac{2 d u}{1 - u^{2} - a - {a u}^{2}} = \int \frac{2 d u}{1 - a - (1 + a) u^{2}}$ $= \int \frac{2 d u}{(1 + a) u^{2} + a - 1} . . .$
Commented by maxmathsup by imad last updated on 26/Apr/19

$\Rightarrow I = \int \frac{2 d u}{(1 + a) u^{2} + a - 1} = \frac{2}{1 + a} \int \frac{d u}{u^{2} + \frac{a - 1}{a + 1}} i f \frac{a - 1}{a + 1} > 0 w e d o t h e c h a n g e m e n t$ $u = \sqrt{\frac{a - 1}{a + 1}} t \Rightarrow I = \frac{2}{1 + a} \frac{a + 1}{a - 1} \int \frac{1}{t^{2} + 1} \sqrt{\frac{a - 1}{a + 1}} d t$ $= \frac{2}{\sqrt{a^{2} - 1}} a r c t a n (\sqrt{\frac{a + 1}{a - 1}} u) + c_{1} = \frac{2}{\sqrt{a^{2} - 1}} a r c t a n (\sqrt{\frac{a + 1}{a - 1}} t a n (\frac{t}{2})) + c_{1} . . . .$
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