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Question Number 32305 by abdo imad last updated on 22/Mar/18

$$find{\int }_1^{e}sin\left(ln\left(x\right)\right)dx.$$

Commented by abdo imad last updated on 24/Mar/18

$$letusethech.lnx=t\Rightarrow I={\int }_0^{1}sint{e}^{t}dtand$$$$I=Im\left({\int }_0^1{e}^{it+t}dt\right)=Im\left({\int }_0^1{e}^{(1+i)t}dt\right)but$$$${\int }_0^1{e}^{(1+i)t}dt=\frac{1}{1+i}{\left[e^{(1+i)t}\right]}_0^1=\frac{1}{1+i}(e^{1+i}-1)$$$$=\frac{1-i}{2}(e(cos\left(1\right)+isin\left(1\right)-1)$$$$=\frac{ecos\left(1\right)+iesin\left(1\right)-1-iecos\left(1\right)+sin\left(1\right)+i}{2}$$$$=\frac{ecos\left(1\right)+sin\left(1\right)-1+i(esin\left(1\right)-ecos\left(1\right)+1)}{2}$$$$I=\frac{1}{2}(esin\left(1\right)-ecos\left(1\right)+1).$$

Answered by sma3l2996 last updated on 23/Mar/18

$$u=sin\left(lnx\right)\Rightarrow u^{\prime }=\frac{1}{x}cos\left(lnx\right)$$$$v^{\prime }=1\Rightarrow v=x$$$$so{\int }_1^{e}sin\left(lnx\right)dx={\left[xsin\left(lnx\right)\right]}_1^e-{\int }_1^{e}cos\left(lnx\right)dx$$$$=esin\left(1\right)-{\int }_1^{e}cos\left(lnx\right)dx$$$$u=cos\left(lnx\right)\Rightarrow u^{\prime }=-\frac{1}{x}sin\left(lnx\right)$$$$v^{\prime }=1\Rightarrow v=x$$$${\int }_1^{e}sin\left(lnx\right)dx=e.sin\left(1\right)-{\left[xcos\left(lnx\right)\right]}_1^e-{\int }_1^{e}sin\left(lnx\right)dx$$$$2{\int }_1^{e}sin\left(lnx\right)dx=e.sin\left(1\right)-e.cos\left(1\right)+1$$$${\int }_1^{e}sin\left(lnx\right)dx=\frac{e}{2}(sin1-cos1)+\frac{1}{2}$$

Commented by abdo imad last updated on 24/Mar/18

$$correctanswerthanks...$$

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