Find-without-any-software-x-5-x-1-5-x-2-sin-ln-x-5-x-dx-

Question Number 173500 by Shrinava last updated on 12/Jul/22

Find without any software :

Ω = \int (x + \frac{5}{x}) (1 - \frac{5}{x^{2}}) \sin (\ln (x + \frac{5}{x})) dx

Answered by thfchristopher last updated on 12/Jul/22

Let u = x + \frac{5}{x}

d u = (1 - \frac{5}{x^{2}}) d x

∴ Ω = \int u \sin (\ln u) d u

= \frac{1}{2} \int \sin (\ln u) d (u^{2})

= \frac{1}{2} [u^{2} \sin (\ln u) - \int u^{2} d {\sin (\ln u)}]

= \frac{1}{2} u^{2} \sin (\ln u) - \frac{1}{2} \int u \cos (\ln u) d u

= \frac{1}{2} u^{2} \sin (\ln u) - \frac{1}{4} \int \cos (\ln u) d (u^{2})

= \frac{1}{2} u^{2} \sin (\ln u) - \frac{1}{4} [u^{2} \cos (\ln u) - \int u^{2} d {\cos (\ln u)}]

= \frac{1}{2} u^{2} \sin (\ln u) - \frac{1}{4} u^{2} \cos (\ln u) - \frac{1}{4} \int u \sin (\ln u) d u

\Rightarrow \frac{5}{4} \int u \sin (\ln u) d u = \frac{1}{4} u^{2} [2 \sin (\ln u) - \cos (\ln u)] + C

\int u \sin (\ln u) d u = \frac{1}{5} u^{2} [2 \sin (\ln u) - \cos (\ln u)] + C

\Rightarrow \int (x + \frac{5}{x}) (1 - \frac{5}{x^{2}}) \sin [\ln (x + \frac{5}{x})] d x

= \frac{1}{5} {(x + \frac{5}{x})}^{2} [2 \sin {\ln (x + \frac{5}{x})} - \cos {\ln (x + \frac{5}{x})}] + C

Commented by Tawa11 last updated on 13/Jul/22

Great sir .

Answered by a.lgnaoui last updated on 13/Jul/22

1 - \frac{5}{x^{2}} = {(x + \frac{5}{x})}^{'} a n d \frac{{(x + \frac{5}{x})}^{'}}{x + \frac{5}{x}} = \ln {(x + \frac{5}{x})}^{'}

⌋ Ω = \int {(x + \frac{5}{x})}^{2} \times \ln {(x + \frac{5}{x})}^{'} \sin (\ln (x + \frac{5}{x})) dx

= \int {(x + \frac{5}{x})}^{2} \times [- \cos {(\ln (x + \frac{5}{x})]}^{'} d x

= l e t U = {(x + \frac{5}{x})}^{2} a n d V = - \cos (\ln (x + \frac{5}{x}))

Ω = \int U \times V^{'} = UV - \int U^{'} V

[- {(x + \frac{5}{x})}^{2} \times \cos (\ln (x + \frac{5}{x}))] + \int {[{(x + \frac{5}{x})}^{2}]}^{'} \cos (\ln (x + \frac{5}{x})) dx

Prime causes double exponent: use braces to clarify

p o s o n s \int U^{'} \cos (\ln (x + \frac{5}{x})) = 2 \int (x + \frac{5}{x}) (1 - \frac{5}{x^{2}}) \cos (\ln (x + \frac{5}{x})) dx

2 \int {(x + \frac{5}{x})}^{2} {[\sin (\ln (x + \frac{5}{x}))]}^{^{'}} dx

= - 2 I o r I = [- \cos (\ln (x + \frac{5}{x})) {(x + \frac{5}{x})}^{2}] - 2 I

3 I = - {(x + \frac{5}{x})}^{2} \cos (\ln (x + \frac{5}{x})) \Rightarrow

I = - \frac{[{(x + \frac{5}{x})}^{2} \cos (\ln (x + \frac{5}{x}))]}{3}

Commented by Tawa11 last updated on 13/Jul/22

Great sir

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