Question Number 27073 by sorour87 last updated on 01/Jan/18
$$\int\mathrm{ln}\:{x}×\mathrm{cos}\:\mathrm{2ln}\:{xdx} \\ $$
Answered by prakash jain last updated on 02/Jan/18
![u=ln x du=(1/x)dx⇒dx=xdu⇒dx=e^u du =∫ue^u cos 2udu cos 2u=((e^(i2u) +e^(−i2u) )/2) (i=(√(−1))) =∫u((e^((1+2i)u) +e^((1−2i)u) )/2)du ∫ue^((1+2i)u) du u(e^((1+2i)u) /(1+2i))−∫(e^((1+2i)u) /(1+2i))du u(e^((1+2i)u) /(1+2i))−(e^((1+2i)u) /((1+2i)^2 )) total integral (1/2)[u(e^((1+2i)u) /(1+2i))−(e^((1+2i)u) /((1+2i)^2 ))+u(e^((1−2i)u) /(1−2i))−(e^((1−2i)u) /((1w−2i)^2 ))] please recheck answer to get the answer in cos/sin substitute e^(iθ) =cos θ+isin θ](https://www.tinkutara.com/question/Q27160.png)
$${u}=\mathrm{ln}\:{x} \\ $$$${du}=\frac{\mathrm{1}}{{x}}{dx}\Rightarrow{dx}={xdu}\Rightarrow{dx}={e}^{{u}} {du} \\ $$$$=\int{ue}^{{u}} \mathrm{cos}\:\mathrm{2}{udu} \\ $$$$\mathrm{cos}\:\mathrm{2}{u}=\frac{{e}^{{i}\mathrm{2}{u}} +{e}^{−{i}\mathrm{2}{u}} }{\mathrm{2}}\:\:\left({i}=\sqrt{−\mathrm{1}}\right) \\ $$$$=\int{u}\frac{{e}^{\left(\mathrm{1}+\mathrm{2}{i}\right){u}} +{e}^{\left(\mathrm{1}−\mathrm{2}{i}\right){u}} }{\mathrm{2}}{du} \\ $$$$\int{ue}^{\left(\mathrm{1}+\mathrm{2}{i}\right){u}} {du} \\ $$$${u}\frac{{e}^{\left(\mathrm{1}+\mathrm{2}{i}\right){u}} }{\mathrm{1}+\mathrm{2}{i}}−\int\frac{{e}^{\left(\mathrm{1}+\mathrm{2}{i}\right){u}} }{\mathrm{1}+\mathrm{2}{i}}{du} \\ $$$${u}\frac{{e}^{\left(\mathrm{1}+\mathrm{2}{i}\right){u}} }{\mathrm{1}+\mathrm{2}{i}}−\frac{{e}^{\left(\mathrm{1}+\mathrm{2}{i}\right){u}} }{\left(\mathrm{1}+\mathrm{2}{i}\right)^{\mathrm{2}} } \\ $$$${total}\:{integral} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[{u}\frac{{e}^{\left(\mathrm{1}+\mathrm{2}{i}\right){u}} }{\mathrm{1}+\mathrm{2}{i}}−\frac{{e}^{\left(\mathrm{1}+\mathrm{2}{i}\right){u}} }{\left(\mathrm{1}+\mathrm{2}{i}\right)^{\mathrm{2}} }+{u}\frac{{e}^{\left(\mathrm{1}−\mathrm{2}{i}\right){u}} }{\mathrm{1}−\mathrm{2}{i}}−\frac{{e}^{\left(\mathrm{1}−\mathrm{2}{i}\right){u}} }{\left(\mathrm{1}{w}−\mathrm{2}{i}\right)^{\mathrm{2}} }\right] \\ $$$$\mathrm{please}\:\mathrm{recheck}\:\mathrm{answer} \\ $$$$\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{in}\:\mathrm{cos}/\mathrm{sin} \\ $$$$\mathrm{substitute}\:{e}^{{i}\theta} =\mathrm{cos}\:\theta+{i}\mathrm{sin}\:\theta \\ $$