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Question Number 156914 by cortano last updated on 17/Oct/21

	Answered by puissant last updated on 17/Oct/21
	$D=∫((arcsin((1/x)))/x^5 )dx ; u=(1/x) → du=−(1/x^2 )dx ⇒ D=−∫ ((u^5 arcsin(u))/u^2 )du=−∫u^3 arcsin(u)du IBP⇒ D=−3[u^2 arcsin(u)]+3∫(u^2 /( (√(1−u^2 ))))du ⇒ D=−3u^2 arcsin(u)−3∫((1−u^2 −1)/( (√(1−u^2 ))))du ⇒ D=−3u^2 arcsin(u)−3∫(√(1−u^2 ))+3∫(1/( (√(1−u^2 ))))du ⇒ D=−3u^2 arcsin(u)−(3/2)[t+(1/2)sin2t]+3arcsin(u)+C ⇒ D=((−3)/x^2 )arcsin((1/x))−(3/2)arcsin((1/x))−(3/2)sin(2arcsin((1/x)))+3arcsin((1/x))+C ∴∵ D=3{((1/2)−(1/x^2 ))arcsin((1/x))−(1/2)sin(2arcsin((1/x)))}+C..$
	$D = \int \frac{a r c s i n (\frac{1}{x})}{x^{5}} d x; u = \frac{1}{x} \to d u = - \frac{1}{x^{2}} d x$ $\Rightarrow D = - \int \frac{u^{5} a r c s i n (u)}{u^{2}} d u = - \int u^{3} a r c s i n (u) d u$ $I B P \Rightarrow D = - 3 [u^{2} a r c s i n (u)] + 3 \int \frac{u^{2}}{\sqrt{1 - u^{2}}} d u$ $\Rightarrow D = - 3 u^{2} a r c s i n (u) - 3 \int \frac{1 - u^{2} - 1}{\sqrt{1 - u^{2}}} d u$ $\Rightarrow D = - 3 u^{2} a r c s i n (u) - 3 \int \sqrt{1 - u^{2}} + 3 \int \frac{1}{\sqrt{1 - u^{2}}} d u$ $\Rightarrow D = - 3 u^{2} a r c s i n (u) - \frac{3}{2} [t + \frac{1}{2} s i n 2 t] + 3 a r c s i n (u) + C$ $\Rightarrow D = \frac{- 3}{x^{2}} a r c s i n (\frac{1}{x}) - \frac{3}{2} a r c s i n (\frac{1}{x}) - \frac{3}{2} s i n (2 a r c s i n (\frac{1}{x})) + 3 a r c s i n (\frac{1}{x}) + C$ $∴∵ D = 3 {(\frac{1}{2} - \frac{1}{x^{2}}) a r c s i n (\frac{1}{x}) - \frac{1}{2} s i n (2 a r c s i n (\frac{1}{x}))} + C . .$
	Commented by cortano last updated on 17/Oct/21

	$y e s .$
	Answered by cortano last updated on 17/Oct/21

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