∫cos^3 xsin^3 xdx


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Question Number 75093 by Rio Michael last updated on 07/Dec/19

$\int {c o s}^{3} {x s i n}^{3} x d x$
Commented by mathmax by abdo last updated on 07/Dec/19
$I =∫ cos^3 x sin^3 xdx =∫ (cosxsinx)^3 dx =(1/8) ∫ (sin(2x))^3 dx =(1/8) ∫ sin(2x)sin^2 (2x)dx =(1/8) ∫ sin(2x)(((1−cos(4x))/2))dx =(1/(16)) ∫ sin(2x)dx−(1/(16)) ∫ sin(2x)cos(4x)dx sin(2x)cos(4x) =cos((π/2)−2x)cos(4x) =(1/2){ cos((π/2)+2x) +cos((π/2)−6x)} =(1/2){sin(6x)−sin(2x)} ⇒ I =−(1/(32)) cos(2x)−(1/(32))∫( sin(6x)−sin(2x))dx =−(1/(32))cos(2x)+(1/(6×32)) cos(6x)−(1/(64)) cos(2x)+C =−(3/(64)) cos(2x) +(1/(192)) cos(6x) +C$
$I = \int {c o s}^{3} x {s i n}^{3} x d x = \int {(c o s x s i n x)}^{3} d x$ $= \frac{1}{8} \int {(s i n (2 x))}^{3} d x = \frac{1}{8} \int s i n (2 x) {s i n}^{2} (2 x) d x$ $= \frac{1}{8} \int s i n (2 x) (\frac{1 - c o s (4 x)}{2}) d x$ $= \frac{1}{16} \int s i n (2 x) d x - \frac{1}{16} \int s i n (2 x) c o s (4 x) d x$ $s i n (2 x) c o s (4 x) = c o s (\frac{π}{2} - 2 x) c o s (4 x)$ $= \frac{1}{2} {c o s (\frac{π}{2} + 2 x) + c o s (\frac{π}{2} - 6 x)} = \frac{1}{2} {s i n (6 x) - s i n (2 x)} \Rightarrow$ $I = - \frac{1}{32} c o s (2 x) - \frac{1}{32} \int (s i n (6 x) - s i n (2 x)) d x$ $= - \frac{1}{32} c o s (2 x) + \frac{1}{6 \times 32} c o s (6 x) - \frac{1}{64} c o s (2 x) + C$ $= - \frac{3}{64} c o s (2 x) + \frac{1}{192} c o s (6 x) + C$
	Answered by mr W last updated on 07/Dec/19

	$\int {c o s}^{3} {x s i n}^{3} x d x$ $= \int {c o s}^{2} {x s i n}^{3} x d (\sin x)$ $= \int (1 - \sin^{2} x) {s i n}^{3} x d (\sin x)$ $= \int (\sin^{3} x - \sin^{5} x) d (\sin x)$ $= \frac{\sin^{4} x}{4} - \frac{\sin^{6} x}{6} + C$
	Commented by Rio Michael last updated on 07/Dec/19

	$t h a n k y o u s i r, s o m u c h$
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