Question-156914

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

Answered by puissant last updated on 17/Oct/21

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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