Question Number 152778 by mnjuly1970 last updated on 01/Sep/21
![Solve .......... Ω := ∫_0 ^( 1) x. sin( ln (x ))dx =^? ((−1)/( 5)) solution.... Ω :=^(i.b.p) [ (x^( 2) /2) . sin(ln(x))]_0 ^( 1) −(1/2)∫_0 ^( 1) x.cos( ln (x ))dx := ((−1)/2) ∫_0 ^( 1) x. cos (ln (x ))dx :=^(i.b.p) ((−1)/2) {[ (x^( 2) /2) cos (ln(x ))]_0 ^1 +(1/2) ∫x. sin(ln(x ))dx} := ((−1)/4) −(1/4) Ω (5/4) Ω = ((−1)/4) ⇒ Ω := ((−1)/( 5)) .........■ m.n .................................](https://www.tinkutara.com/question/Q152778.png)
$$ \\ $$$$\:\:\:\:\:\:\:\:\mathrm{Solve}\:………. \\ $$$$\:\:\:\:\Omega\::=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}.\:{sin}\left(\:{ln}\:\left({x}\:\right)\right){dx}\:\overset{?} {=}\:\frac{−\mathrm{1}}{\:\:\:\mathrm{5}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{solution}…. \\ $$$$\:\:\:\:\:\Omega\::\overset{{i}.{b}.{p}} {=}\left[\:\frac{{x}^{\:\mathrm{2}} }{\mathrm{2}}\:.\:{sin}\left({ln}\left({x}\right)\right)\right]_{\mathrm{0}} ^{\:\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \:{x}.{cos}\left(\:{ln}\:\left({x}\:\right)\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:\::=\:\frac{−\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}.\:{cos}\:\left({ln}\:\left({x}\:\right)\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:\::\overset{{i}.{b}.{p}} {=}\:\frac{−\mathrm{1}}{\mathrm{2}}\:\left\{\left[\:\frac{{x}^{\:\mathrm{2}} }{\mathrm{2}}\:{cos}\:\left({ln}\left({x}\:\right)\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}\:\int{x}.\:{sin}\left({ln}\left({x}\:\right)\right){dx}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\::=\:\frac{−\mathrm{1}}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{4}}\:\Omega \\ $$$$\:\:\:\:\:\:\:\:\:\frac{\mathrm{5}}{\mathrm{4}}\:\Omega\:=\:\frac{−\mathrm{1}}{\mathrm{4}}\:\:\Rightarrow\:\:\:\Omega\::=\:\frac{−\mathrm{1}}{\:\mathrm{5}}\:\:………\blacksquare\:{m}.{n} \\ $$$$\:\:\:\:……………………………\:\: \\ $$$$\:\:\:\:\:\:\: \\ $$