Question Number 151 by 123456 last updated on 25/Jan/15 | ||
$$\mathrm{evaluate}\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{sin}\:{x}}{{x}}\mathrm{ln}\:{x}\:{dx} \\ $$ | ||
Answered by prakash jain last updated on 13/Dec/14 | ||
$$\mathrm{sin}\:{x}={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}−\frac{{x}^{\mathrm{7}} }{\mathrm{7}!}+.... \\ $$$$\frac{\mathrm{sin}\:{x}}{{x}}=\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{5}!}−\frac{{x}^{\mathrm{6}} }{\mathrm{7}!}+.... \\ $$$$\mathrm{Integrating}\:\int{uvdx},\:{u}=\mathrm{ln}\:{x},\:{v}=\frac{\mathrm{sin}\:{x}}{{x}} \\ $$$$\mathrm{ln}\:{x}\centerdot\left[{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}\centerdot\mathrm{3}!}+...\right]−\int\frac{\mathrm{1}}{{x}}\centerdot\left[{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}\centerdot\mathrm{3}!}+...\right]{dx} \\ $$$$\mathrm{ln}\:{x}\centerdot\left[{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}\centerdot\mathrm{3}!}+...\right]−\int\left[\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{3}\centerdot\mathrm{3}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{5}\centerdot\mathrm{5}!}−\frac{{x}^{\mathrm{6}} }{\mathrm{7}\centerdot\mathrm{7}!}...\right]{dx} \\ $$$$\mathrm{ln}\:{x}\centerdot\left[{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}\centerdot\mathrm{3}!}+...\right]−\left[{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{2}} \centerdot\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}^{\mathrm{2}} \centerdot\mathrm{5}!}−\frac{{x}^{\mathrm{7}} }{\mathrm{7}^{\mathrm{2}} \centerdot\mathrm{7}!}...\right] \\ $$$$\mathrm{integrating}\:\mathrm{from}\:\mathrm{0}\:\mathrm{to}\:\mathrm{1}\:\mathrm{first}\:\mathrm{part}\:\mathrm{is}\:\mathrm{0} \\ $$$$\mathrm{result}=−\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{54}}+\frac{\mathrm{1}}{\mathrm{3000}}−\frac{\mathrm{1}}{\mathrm{246960}}+...\right] \\ $$$$\approx−\mathrm{0}.\mathrm{98181} \\ $$ | ||