Question Number 173500 by Shrinava last updated on 12/Jul/22
$$\mathrm{Find}\:\mathrm{without}\:\mathrm{any}\:\mathrm{software}: \\ $$$$\Omega\:=\:\int\:\left(\mathrm{x}\:+\:\frac{\mathrm{5}}{\mathrm{x}}\right)\left(\mathrm{1}\:−\:\frac{\mathrm{5}}{\mathrm{x}^{\mathrm{2}} }\right)\mathrm{sin}\left(\mathrm{ln}\left(\mathrm{x}\:+\:\frac{\mathrm{5}}{\mathrm{x}}\right)\right)\mathrm{dx} \\ $$
Answered by thfchristopher last updated on 12/Jul/22
![Let u=x+(5/x) du=(1−(5/x^2 ))dx ∴ Ω=∫usin (ln u)du =(1/2)∫sin (ln u)d(u^2 ) =(1/2)[u^2 sin (ln u)−∫u^2 d{sin (ln u)}] =(1/2)u^2 sin (ln u)−(1/2)∫ucos (ln u)du =(1/2)u^2 sin (ln u)−(1/4)∫cos (ln u)d(u^2 ) =(1/2)u^2 sin (ln u)−(1/4)[u^2 cos (ln u)−∫u^2 d{cos (ln u)}] =(1/2)u^2 sin (ln u)−(1/4)u^2 cos (ln u)−(1/4)∫usin (ln u)du ⇒(5/4)∫usin (ln u)du=(1/4)u^2 [2sin (ln u)−cos (ln u)]+C ∫usin (ln u)du=(1/5)u^2 [2sin (ln u)−cos (ln u)]+C ⇒∫(x+(5/x))(1−(5/x^2 ))sin [ln (x+(5/x))]dx =(1/5)(x+(5/x))^2 [2sin {ln (x+(5/x))}−cos {ln (x+(5/x))}]+C](Q173514.png)
$$\mathrm{Let}\:{u}={x}+\frac{\mathrm{5}}{{x}} \\ $$$${du}=\left(\mathrm{1}−\frac{\mathrm{5}}{{x}^{\mathrm{2}} }\right){dx} \\ $$$$\therefore\:\Omega=\int{u}\mathrm{sin}\:\left(\mathrm{ln}\:{u}\right){du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{sin}\:\left(\mathrm{ln}\:{u}\right){d}\left({u}^{\mathrm{2}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{u}^{\mathrm{2}} \mathrm{sin}\:\left(\mathrm{ln}\:{u}\right)−\int{u}^{\mathrm{2}} {d}\left\{\mathrm{sin}\:\left(\mathrm{ln}\:{u}\right)\right\}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{u}^{\mathrm{2}} \mathrm{sin}\:\left(\mathrm{ln}\:{u}\right)−\frac{\mathrm{1}}{\mathrm{2}}\int{u}\mathrm{cos}\:\left(\mathrm{ln}\:{u}\right){du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{u}^{\mathrm{2}} \mathrm{sin}\:\left(\mathrm{ln}\:{u}\right)−\frac{\mathrm{1}}{\mathrm{4}}\int\mathrm{cos}\:\left(\mathrm{ln}\:{u}\right){d}\left({u}^{\mathrm{2}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{u}^{\mathrm{2}} \mathrm{sin}\:\left(\mathrm{ln}\:{u}\right)−\frac{\mathrm{1}}{\mathrm{4}}\left[{u}^{\mathrm{2}} \mathrm{cos}\:\left(\mathrm{ln}\:{u}\right)−\int{u}^{\mathrm{2}} {d}\left\{\mathrm{cos}\:\left(\mathrm{ln}\:{u}\right)\right\}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{u}^{\mathrm{2}} \mathrm{sin}\:\left(\mathrm{ln}\:{u}\right)−\frac{\mathrm{1}}{\mathrm{4}}{u}^{\mathrm{2}} \mathrm{cos}\:\left(\mathrm{ln}\:{u}\right)−\frac{\mathrm{1}}{\mathrm{4}}\int{u}\mathrm{sin}\:\left(\mathrm{ln}\:{u}\right){du} \\ $$$$\Rightarrow\frac{\mathrm{5}}{\mathrm{4}}\int{u}\mathrm{sin}\:\left(\mathrm{ln}\:{u}\right){du}=\frac{\mathrm{1}}{\mathrm{4}}{u}^{\mathrm{2}} \left[\mathrm{2sin}\:\left(\mathrm{ln}\:{u}\right)−\mathrm{cos}\:\left(\mathrm{ln}\:{u}\right)\right]+{C} \\ $$$$\int{u}\mathrm{sin}\:\left(\mathrm{ln}\:{u}\right){du}=\frac{\mathrm{1}}{\mathrm{5}}{u}^{\mathrm{2}} \left[\mathrm{2sin}\:\left(\mathrm{ln}\:{u}\right)−\mathrm{cos}\:\left(\mathrm{ln}\:{u}\right)\right]+{C} \\ $$$$\Rightarrow\int\left({x}+\frac{\mathrm{5}}{{x}}\right)\left(\mathrm{1}−\frac{\mathrm{5}}{{x}^{\mathrm{2}} }\right)\mathrm{sin}\:\left[\mathrm{ln}\:\left({x}+\frac{\mathrm{5}}{{x}}\right)\right]{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\left({x}+\frac{\mathrm{5}}{{x}}\right)^{\mathrm{2}} \left[\mathrm{2sin}\:\left\{\mathrm{ln}\:\left({x}+\frac{\mathrm{5}}{{x}}\right)\right\}−\mathrm{cos}\:\left\{\mathrm{ln}\:\left({x}+\frac{\mathrm{5}}{{x}}\right)\right\}\right]+{C} \\ $$
Commented by Tawa11 last updated on 13/Jul/22
$$\mathrm{Great}\:\mathrm{sir}. \\ $$
Answered by a.lgnaoui last updated on 13/Jul/22
![1−(5/x^2 )=(x+(5/x))′ and (((x+(5/x))′)/(x+(5/x)))=ln(x+(5/x))′ ⌋Ω=∫(x+(5/x))^2 ×ln(x+(5/x))′sin (ln(x+(5/x)))dx =∫(x+(5/x))^2 ×[−cos (ln(x+(5/x))]′dx =let U=(x+(5/x))^2 and V=−cos (ln(x+(5/x))) Ω=∫U×V′ =UV−∫U′V [−(x+(5/x))^2 ×cos (ln(x+(5/x))) ]+∫[(x+(5/x))^2 ]′cos (ln(x+(5/x)))dx (x+(5/x))^2 ′=2(x+(5/x))(1−(5/x^2 )) posons ∫U′cos (ln(x+(5/x)))=2∫(x+(5/x))(1−(5/x^2 ))cos (ln(x+(5/x)))dx 2∫(x+(5/x))^2 [sin( ln(x+(5/x)))]^′ dx =−2I or I=[−cos(ln (x+(5/x)))(x+(5/x))^2 ]−2I 3I=−(x+(5/x))^2 cos (ln(x+(5/x)))⇒ I= −(([(x+(5/x))^2 cos (ln(x+(5/x)))])/3)](Q173524.png)
$$\mathrm{1}−\frac{\mathrm{5}}{\mathrm{x}^{\mathrm{2}} }=\left(\mathrm{x}+\frac{\mathrm{5}}{\mathrm{x}}\right)'\:\:{and}\:\:\:\frac{\left(\mathrm{x}+\frac{\mathrm{5}}{\mathrm{x}}\right)'}{{x}+\frac{\mathrm{5}}{\mathrm{x}}}=\mathrm{ln}\left(\mathrm{x}+\frac{\mathrm{5}}{\mathrm{x}}\right)' \\ $$$$\rfloor\Omega=\int\left({x}+\frac{\mathrm{5}}{{x}}\right)^{\mathrm{2}} ×\mathrm{ln}\left({x}+\frac{\mathrm{5}}{{x}}\right)'\mathrm{sin}\:\left(\mathrm{ln}\left(\mathrm{x}+\frac{\mathrm{5}}{\mathrm{x}}\right)\right)\mathrm{dx} \\ $$$$=\int\left({x}+\frac{\mathrm{5}}{{x}}\right)^{\mathrm{2}} ×\left[−\mathrm{cos}\:\left(\mathrm{ln}\left(\mathrm{x}+\frac{\mathrm{5}}{\mathrm{x}}\right)\right]'{dx}\right. \\ $$$$={let}\:\:\:\:\mathrm{U}=\left({x}+\frac{\mathrm{5}}{{x}}\right)^{\mathrm{2}} \:\:\:\:{and}\:\:\mathrm{V}=−\mathrm{cos}\:\left(\mathrm{ln}\left(\mathrm{x}+\frac{\mathrm{5}}{\mathrm{x}}\right)\right) \\ $$$$\Omega=\int\mathrm{U}×\mathrm{V}'\:\:=\mathrm{UV}−\int\mathrm{U}'\mathrm{V} \\ $$$$\left[−\left({x}+\frac{\mathrm{5}}{{x}}\right)^{\mathrm{2}} ×\mathrm{cos}\:\left(\mathrm{ln}\left(\mathrm{x}+\frac{\mathrm{5}}{\mathrm{x}}\right)\right)\:\right]+\int\left[\left({x}+\frac{\mathrm{5}}{{x}}\right)^{\mathrm{2}} \right]'\mathrm{cos}\:\left(\mathrm{ln}\left(\mathrm{x}+\frac{\mathrm{5}}{\mathrm{x}}\right)\right)\mathrm{dx} \\ $$$$\left(\mathrm{x}+\frac{\mathrm{5}}{\mathrm{x}}\right)^{\mathrm{2}} '=\mathrm{2}\left({x}+\frac{\mathrm{5}}{{x}}\right)\left(\mathrm{1}−\frac{\mathrm{5}}{{x}^{\mathrm{2}} }\right) \\ $$$${posons}\:\:\int\mathrm{U}'\mathrm{cos}\:\left(\mathrm{ln}\left(\mathrm{x}+\frac{\mathrm{5}}{\mathrm{x}}\right)\right)=\mathrm{2}\int\left({x}+\frac{\mathrm{5}}{\mathrm{x}}\right)\left(\mathrm{1}−\frac{\mathrm{5}}{{x}^{\mathrm{2}} }\right)\mathrm{cos}\:\left(\mathrm{ln}\left(\mathrm{x}+\frac{\mathrm{5}}{\mathrm{x}}\right)\right)\mathrm{dx} \\ $$$$ \\ $$$$\mathrm{2}\int\left({x}+\frac{\mathrm{5}}{{x}}\right)^{\mathrm{2}} \left[\mathrm{sin}\left(\:\mathrm{ln}\left(\mathrm{x}+\frac{\mathrm{5}}{\mathrm{x}}\right)\right)\right]^{'} \mathrm{dx} \\ $$$$=−\mathrm{2I}\:\:\:\:{or}\:\:\mathrm{I}=\left[−\mathrm{cos}\left(\mathrm{ln}\:\left({x}+\frac{\mathrm{5}}{{x}}\right)\right)\left(\mathrm{x}+\frac{\mathrm{5}}{\mathrm{x}}\right)^{\mathrm{2}} \right]−\mathrm{2I} \\ $$$$\mathrm{3I}=−\left({x}+\frac{\mathrm{5}}{{x}}\right)^{\mathrm{2}} \mathrm{cos}\:\left(\mathrm{ln}\left(\mathrm{x}+\frac{\mathrm{5}}{\mathrm{x}}\right)\right)\Rightarrow \\ $$$$\mathrm{I}=\:−\frac{\left[\left({x}+\frac{\mathrm{5}}{{x}}\right)^{\mathrm{2}} \mathrm{cos}\:\left(\mathrm{ln}\left(\mathrm{x}+\frac{\mathrm{5}}{\mathrm{x}}\right)\right)\right]}{\mathrm{3}} \\ $$
Commented by Tawa11 last updated on 13/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$