1) ∫_0 ^(Π/2) sin x∙sin 2x∙sin 3x∙dx = ? 2) ∫


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Question Number 109495 by qwerty111 last updated on 24/Aug/20

$1) \int_{0}^{\frac{Π}{2}} \sin x \cdot \sin 2 x \cdot \sin 3 x \cdot d x = ?$ $2) \int_{0}^{\frac{1}{2}} \arcsin x \cdot d x = ?$
	Answered by bemath last updated on 24/Aug/20
	$1) sin 3x = 3sin x−4sin^3 x I=∫_0 ^(π/2) 2sin^2 x.(3sin x−4sin^3 x)cos x dx= set u = sin x → { ((u=1)),((u=0)) :} I=∫_0 ^1 2u^2 (3u−4u^3 ) du = I= ∫_0 ^1 (6u^3 −8u^5 )du = [(3/2)u^4 −(4/3)u^6 ]_0 ^1 = (3/2)−(4/3) = (1/6)$
	$1) \sin 3 x = 3 \sin x - 4 \sin^{3} x$ $I = {\int^{π / 2}}_{0} {2 \sin}^{2} x . (3 \sin x - 4 \sin^{3} x) \cos x d x =$ $s e t u = \sin x \to {\begin{cases} u = 1 \\ u = 0 \end{cases}$ $I = \underset{0}{\int^{1}} 2 u^{2} (3 u - 4 u^{3}) d u =$ $I = \underset{0}{\int^{1}} (6 u^{3} - 8 u^{5}) d u = {[\frac{3}{2} u^{4} - \frac{4}{3} u^{6}]}_{0}^{1}$ $= \frac{3}{2} - \frac{4}{3} = \frac{1}{6}$
	Answered by bemath last updated on 24/Aug/20
	$(2) by part { ((u=arc sin x ⇒du = (dx/( (√(1−x^2 )))))),((dv = dx →v = x)) :} I= [ x arcsin (x)]_0 ^(1/2) −∫^(1/2) _0 ((x dx)/( (√(1−x^2 )))) I= (1/2).(π/6)+(1/2)∫_0 ^(1/2) ((d(1−x^2 ))/( (√(1−x^2 )))) I=(π/(12))+[ (√(1−x^2 )) ]_0 ^(1/2) I=(π/(12)) + (((√3)/2)−1) = ((π+6(√3)−12)/(12))$
	$(2) b y p a r t {\begin{cases} u = arc \sin x \Rightarrow d u = \frac{d x}{\sqrt{1 - x^{2}}} \\ d v = d x \to v = x \end{cases}$ $I = {[x \arcsin (x)]}_{0}^{1 / 2} - {\int_{0}}^{1 / 2} \frac{x d x}{\sqrt{1 - x^{2}}}$ $I = \frac{1}{2} . \frac{π}{6} + \frac{1}{2} {\int^{1 / 2}}_{0} \frac{d (1 - x^{2})}{\sqrt{1 - x^{2}}}$ $I = \frac{π}{12} + {[\sqrt{1 - x^{2}}]}_{0}^{1 / 2}$ $I = \frac{π}{12} + (\frac{\sqrt{3}}{2} - 1) = \frac{π + 6 \sqrt{3} - 12}{12}$
	Answered by 1549442205PVT last updated on 24/Aug/20

	$1) sinx . \sin 3 x = \frac{\cos 2 x - \cos 4 x}{2} . Hence,$ $F = \int_{0}^{\frac{Π}{2}} \sin x \cdot \sin 2 x \cdot \sin 3 x \cdot d x =$ $= \frac{1}{2} \int_{0}^{\frac{π}{2}} \sin 2 xcos 2 xdx - \frac{1}{2} \int_{0}^{\frac{π}{2}} \sin 2 xcos 4 xdx$ $= \frac{1}{4} \int_{0}^{\frac{π}{2}} \sin 4 xdx - \frac{1}{2} \int_{0}^{\frac{π}{2}} \sin 2 x ({2 \cos}^{2} x - 1)$ $= - \frac{1}{16} \cos 4 x ∣_{0}^{\frac{π}{2}} + \frac{1}{2} \int_{0}^{\frac{π}{2}} \sin 2 xdx$ $+ \frac{1}{2} \int_{0}^{\frac{π}{2}} \cos^{2} 2 xdcos 2 x$ $= 0 - \frac{1}{4} \cos 2 x ∣_{0}^{\frac{π}{2}} + \frac{1}{6} \cos^{3} 2 x ∣_{0}^{\frac{π}{2}}$ $= 0 - \frac{1}{4} (- 1 - 1) + \frac{1}{6} (- 1 - 1) =$ $= \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$
	Answered by mathmax by abdo last updated on 24/Aug/20

	$2) I = \int_{0}^{\frac{1}{2}} arcsinx dx we do the chsngement arcsinx = t \Rightarrow x = sint$ $\Rightarrow I = \int_{0}^{\frac{π}{6}} t cost dt = {[tsint]}_{0}^{\frac{π}{6}} - \int_{0}^{\frac{π}{6}} sint dt$ $I = \frac{π}{6} . \frac{1}{2} + {[cost]}_{0}^{\frac{π}{6}} = \frac{π}{12} + (\frac{\sqrt{3}}{2} - 1) \Rightarrow I = \frac{π}{12} + \frac{\sqrt{3}}{2} - 1$
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