find ∫_(−(π/3)) ^(π/3) x^2 {cosx−sinx}^3 dx


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Question Number 67235 by prof Abdo imad last updated on 24/Aug/19
$find ∫_(−(π/3)) ^(π/3) x^2 {cosx−sinx}^3 dx$
$f i n d \int_{- \frac{π}{3}}^{\frac{π}{3}} x^{2} {c o s x - s i n x}^{3} d x$
Commented by mathmax by abdo last updated on 01/Sep/19
$let I =∫_(−(π/3)) ^(π/3) x^2 {cosx −sinx}^3 and J=∫_(−(π/6)) ^(π/6) x^2 {cosx +sinx}^3 dx we have I +J =∫_(−(π/3)) ^(π/3) x^2 (cosx −sinx +cosx +sinx)((cosx−sinx)^2 −(cosx−sinx)(cosx +sinx)+(cosx +sinx)^2 }dx =∫_(−(π/3)) ^(π/3) 2x^2 cosx{1−2cosxsinx −(cos^2 x−sin^2 x) +1+2cosx sinx}dx =∫_(−(π/3)) ^(π/3) 2x^2 cosx{2 −cos(2x)}dx =4 ∫_0 ^(π/3) x^2 cosx(2−cos(2x))dx =8 ∫_0 ^(π/3) x^2 cosxdx −4∫_0 ^(π/3) cosx.cos(2x)dx by parts ∫_0 ^(π/3) x^2 cosxdx=[x^2 sinx]_0 ^(π/3) −∫_0 ^(π/3) 2xsinx dx =−2{[−xcosx]_0 ^(π/3) −∫_0 ^(π/3) (−cosx)dx} =−2{−(π/6) +[sinx]_0 ^(π/3) }=−2{−(π/6)+((√3)/2)} =(π/3)−(√3) ∫_0 ^(π/3) cosx cos(2x)dx =(1/2)∫_0 ^(π/3) (cos(3x)+cosx)dx=(1/6)[sin(3x)]_0 ^(π/3) +(1/2)[sinx]_0 ^(π/3) =(1/2)((√3)/2) =((√3)/4) ⇒ I+J =8((π/3)−(√3))−(√3) =((8π)/3)−9(√3) I−J =∫_(−(π/3)) ^(π/3) x^2 {(cosx−sinx)^3 −(cosx+sinx)^3 }dx =∫_(−(π/3)) ^(π/3) x^2 (−2sinx)(1−2cosx sinx +cos(2x) +1+2cosxsinx}dx =−2∫_(−(π/3)) ^(π/3) x^2 sinx{2+cos(2x)}dx =0 because the function is odd ⇒I−J =0 ⇒I =J ⇒ ⇒2I =((8π)/3)−9(√3) ⇒I =((4π)/3)−(9/2)(√3)$
$l e t I = \int_{- \frac{π}{3}}^{\frac{π}{3}} x^{2} {c o s x - s i n x}^{3} a n d J = \int_{- \frac{π}{6}}^{\frac{π}{6}} x^{2} {c o s x + s i n x}^{3} d x$ $w e h a v e I + J = \int_{- \frac{π}{3}}^{\frac{π}{3}} x^{2} (c o s x - s i n x + c o s x + s i n x) ({(c o s x - s i n x)}^{2}$ $- (c o s x - s i n x) (c o s x + s i n x) + {(c o s x + s i n x)}^{2}} d x$ $= \int_{- \frac{π}{3}}^{\frac{π}{3}} 2 x^{2} c o s x {1 - 2 c o s x s i n x - ({c o s}^{2} x - {s i n}^{2} x) + 1 + 2 c o s x s i n x} d x$ $= \int_{- \frac{π}{3}}^{\frac{π}{3}} 2 x^{2} c o s x {2 - c o s (2 x)} d x$ $= 4 \int_{0}^{\frac{π}{3}} x^{2} c o s x (2 - c o s (2 x)) d x = 8 \int_{0}^{\frac{π}{3}} x^{2} c o s x d x - 4 \int_{0}^{\frac{π}{3}} c o s x . c o s (2 x) d x$ $b y p a r t s \int_{0}^{\frac{π}{3}} x^{2} c o s x d x = {[x^{2} s i n x]}_{0}^{\frac{π}{3}} - \int_{0}^{\frac{π}{3}} 2 x s i n x d x$ $= - 2 {{[- x c o s x]}_{0}^{\frac{π}{3}} - \int_{0}^{\frac{π}{3}} (- c o s x) d x}$ $= - 2 {- \frac{π}{6} + {[s i n x]}_{0}^{\frac{π}{3}}} = - 2 {- \frac{π}{6} + \frac{\sqrt{3}}{2}} = \frac{π}{3} - \sqrt{3}$ $\int_{0}^{\frac{π}{3}} c o s x c o s (2 x) d x = \frac{1}{2} \int_{0}^{\frac{π}{3}} (c o s (3 x) + c o s x) d x = \frac{1}{6} {[s i n (3 x)]}_{0}^{\frac{π}{3}}$ $+ \frac{1}{2} {[s i n x]}_{0}^{\frac{π}{3}} = \frac{1}{2} \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} \Rightarrow I + J = 8 (\frac{π}{3} - \sqrt{3}) - \sqrt{3} = \frac{8 π}{3} - 9 \sqrt{3}$ $I - J = \int_{- \frac{π}{3}}^{\frac{π}{3}} x^{2} {{(c o s x - s i n x)}^{3} - {(c o s x + s i n x)}^{3}} d x$ $= \int_{- \frac{π}{3}}^{\frac{π}{3}} x^{2} (- 2 s i n x) (1 - 2 c o s x s i n x + c o s (2 x) + 1 + 2 c o s x s i n x} d x$ $= - 2 \int_{- \frac{π}{3}}^{\frac{π}{3}} x^{2} s i n x {2 + c o s (2 x)} d x$ $= 0 b e c a u s e t h e f u n c t i o n i s o d d \Rightarrow I - J = 0 \Rightarrow I = J \Rightarrow$ $\Rightarrow 2 I = \frac{8 π}{3} - 9 \sqrt{3} \Rightarrow I = \frac{4 π}{3} - \frac{9}{2} \sqrt{3}$
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