calculate ∫_(1/2) ^(5/4) (x^3 /(√(2+x−x^2 )))dx


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Question Number 42802 by maxmathsup by imad last updated on 02/Sep/18

$c a l c u l a t e \int_{\frac{1}{2}}^{\frac{5}{4}} \frac{x^{3}}{\sqrt{2 + x - x^{2}}} d x$
Commented by maxmathsup by imad last updated on 05/Sep/18
$let A = ∫_(1/2) ^(5/4) (x^3 /(√(2+x−x^2 )))dx⇒A = ∫_(1/2) ^(5/4) (x^3 /(√(−x^2 +x+2)))dx we have −x^2 +x+2 =−(x^2 −x−2) =−(x^2 −2(1/2)x +(1/4)−(1/4)−2) =−{(x−(1/2))^2 −(9/4)}=(9/4) −(x−(1/2))^2 ⇒A = ∫_(1/2) ^(5/4) (x^3 /(√((9/4)−(x−(1/2))^2 )))dx =_(x−(1/2) =(3/2)sint) ∫_0 ^(π/6) ((((1/2)+(3/2)sint)^3 )/((3/2) cost)) (3/2) cost dt = ∫_0 ^(π/6) (3sint +1)^3 dt =∫_0 ^(π/6) (27sin^3 t +27sin^2 t +9sint +1)dt =27 ∫_0 ^(π/6) sin^3 t dt +27 ∫_0 ^(π/6) sin^2 t dt +9 ∫_0 ^(π/6) sint dt +(π/6) but ∫_0 ^(π/6) sint dt =[−cost]_0 ^(π/6) =1−((√3)/2) ∫_0 ^(π/6) sin^2 t dt =(1/2) ∫_0 ^(π/6) (1−cos(2t))dt=(π/(12)) −(1/4)[sin(2t)]_0 ^(π/6) =(π/(12)) −(1/4) ((√3)/2) =(π/(12)) −((√3)/8) we have sin^3 t ={((e^(it) −e^(−it) )/(2i))}^3 =−(1/(8i)){ Σ_(k=0) ^3 C_3 ^k e^(ikt) (−1)^(3−k) e^(−i(3−k)t) } =(i/8){ −e^(−i3t) +3 e^(it) e^(−i2t) − 3 e^(i2t) e^(−it) +e^(i3t) } =(i/8){e^(i3t) −e^(−i3t) −3(e^(it) −e^(−it) )} =(i/8){2i sin(3t) −6i sin(t)} =−(1/4)sin(3t) +(3/4) sin(t) ⇒ ∫_0 ^(π/6) sin^3 t dt =(3/4) ∫_0 ^(π/6) sin(t)dt−(1/4) ∫_0 ^(π/6) sin(3t)dt =−(3/4)[cost]_0 ^(π/6) +(1/(12))[ cos(3t)]_0 ^(π/6) =−(3/4)(((√3)/2)−1) +(1/(12))( −1) =(3/4) −((3(√3))/8) −(1/(12)) so the value of A is known .$
$l e t A = \int_{\frac{1}{2}}^{\frac{5}{4}} \frac{x^{3}}{\sqrt{2 + x - x^{2}}} d x \Rightarrow A = \int_{\frac{1}{2}}^{\frac{5}{4}} \frac{x^{3}}{\sqrt{- x^{2} + x + 2}} d x$ $w e h a v e - x^{2} + x + 2 = - (x^{2} - x - 2) = - (x^{2} - 2 \frac{1}{2} x + \frac{1}{4} - \frac{1}{4} - 2)$ $= - {{(x - \frac{1}{2})}^{2} - \frac{9}{4}} = \frac{9}{4} - {(x - \frac{1}{2})}^{2} \Rightarrow A = \int_{\frac{1}{2}}^{\frac{5}{4}} \frac{x^{3}}{\sqrt{\frac{9}{4} - {(x - \frac{1}{2})}^{2}}} d x$ $=_{x - \frac{1}{2} = \frac{3}{2} s i n t} \int_{0}^{\frac{π}{6}} \frac{{(\frac{1}{2} + \frac{3}{2} s i n t)}^{3}}{\frac{3}{2} c o s t} \frac{3}{2} c o s t d t$ $= \int_{0}^{\frac{π}{6}} {(3 s i n t + 1)}^{3} d t = \int_{0}^{\frac{π}{6}} (27 {s i n}^{3} t + 27 {s i n}^{2} t + 9 s i n t + 1) d t$ $= 27 \int_{0}^{\frac{π}{6}} {s i n}^{3} t d t + 27 \int_{0}^{\frac{π}{6}} {s i n}^{2} t d t + 9 \int_{0}^{\frac{π}{6}} s i n t d t + \frac{π}{6} b u t$ $\int_{0}^{\frac{π}{6}} s i n t d t = {[- c o s t]}_{0}^{\frac{π}{6}} = 1 - \frac{\sqrt{3}}{2}$ $\int_{0}^{\frac{π}{6}} {s i n}^{2} t d t = \frac{1}{2} \int_{0}^{\frac{π}{6}} (1 - c o s (2 t)) d t = \frac{π}{12} - \frac{1}{4} {[s i n (2 t)]}_{0}^{\frac{π}{6}}$ $= \frac{π}{12} - \frac{1}{4} \frac{\sqrt{3}}{2} = \frac{π}{12} - \frac{\sqrt{3}}{8} w e h a v e {s i n}^{3} t = {\frac{e^{i t} - e^{- i t}}{2 i}}^{3}$ $= - \frac{1}{8 i} {\sum_{k = 0}^{3} C_{3}^{k} e^{i k t} {(- 1)}^{3 - k} e^{- i (3 - k) t}}$ $= \frac{i}{8} {- e^{- i 3 t} + 3 e^{i t} e^{- i 2 t} - 3 e^{i 2 t} e^{- i t} + e^{i 3 t}}$ $= \frac{i}{8} {e^{i 3 t} - e^{- i 3 t} - 3 (e^{i t} - e^{- i t})} = \frac{i}{8} {2 i s i n (3 t) - 6 i s i n (t)}$ $= - \frac{1}{4} s i n (3 t) + \frac{3}{4} s i n (t) \Rightarrow \int_{0}^{\frac{π}{6}} {s i n}^{3} t d t = \frac{3}{4} \int_{0}^{\frac{π}{6}} s i n (t) d t - \frac{1}{4} \int_{0}^{\frac{π}{6}} s i n (3 t) d t$ $= - \frac{3}{4} {[c o s t]}_{0}^{\frac{π}{6}} + \frac{1}{12} {[c o s (3 t)]}_{0}^{\frac{π}{6}} = - \frac{3}{4} (\frac{\sqrt{3}}{2} - 1) + \frac{1}{12} (- 1)$ $= \frac{3}{4} - \frac{3 \sqrt{3}}{8} - \frac{1}{12} s o t h e v a l u e o f A i s k n o w n .$
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