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Question Number 101212 by rb222 last updated on 01/Jul/20

$$\int{e}^{{x}} \mathrm{sin}\:{x}\:{dx}\:=\:−{e}^{{x}} {cos}\:{x}\:+\:{e}^{{x}} \:{sin}\:{x}\:−\:\int\:{e}^{{x}} \:\mathrm{sin}\:\:{x}\:{dx} \\$$

Commented by Dwaipayan Shikari last updated on 01/Jul/20

$$\int{e}^{{x}} {sinxdx}={e}^{{x}} \int{sinxdx}+\int{e}^{{x}} {cosxdx} \\$$$$\:\:\:\:\:\:\:{I}=−{e}^{{x}} {cosx}+{e}^{{x}} {sinx}−\int{e}^{{x}} {sinxdx} \\$$$${I}={e}^{{x}} \left({sinx}−{cosx}\right)−{I} \\$$$$\mathrm{2}{I}={e}^{{x}} \left({sinx}−{cosx}\right)+{C} \\$$$${I}=\frac{{e}^{{x}} \left({sinx}−{cosx}\right)}{\mathrm{2}}+{Constant}\left({or}\:\frac{{C}}{\mathrm{2}}\right) \\$$

Answered by smridha last updated on 01/Jul/20

$$\boldsymbol{{Im}}\int\boldsymbol{{e}}^{\left(\mathrm{1}+\boldsymbol{{i}}\right)\boldsymbol{{x}}} \boldsymbol{{dx}}=\boldsymbol{{I}}{m}\left[\frac{\boldsymbol{{e}}^{\left(\mathrm{1}+\boldsymbol{{i}}\right)\boldsymbol{{x}}} }{\left(\mathrm{1}+\boldsymbol{{i}}\right)}\right]+\boldsymbol{{c}} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{I}}{m}\left[\left(\mathrm{1}−\boldsymbol{{i}}\right)\boldsymbol{{e}}^{\boldsymbol{{x}}} \left(\boldsymbol{{cosx}}+\boldsymbol{{isinx}}\right)\right]+\boldsymbol{{c}} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{I}}{m}\left[\boldsymbol{{e}}^{\boldsymbol{{x}}} \left\{\left(\boldsymbol{{cosx}}+\boldsymbol{{sinx}}\right)+\boldsymbol{{i}}\left(\boldsymbol{{sinx}}−\boldsymbol{{cosx}}\right)\right\}\right]+\boldsymbol{{c}} \\$$$$=\frac{\boldsymbol{{e}}^{\boldsymbol{{x}}} }{\mathrm{2}}\left(\boldsymbol{{sinx}}−\boldsymbol{{cosx}}\right)+\boldsymbol{{c}}\:\:\:\:\:\boldsymbol{{ans}} \\$$$$\boldsymbol{{or}}\:=\frac{\boldsymbol{{e}}^{\boldsymbol{{x}}} }{\sqrt{\mathrm{2}}}\left(\boldsymbol{{sinxcos}}\frac{\boldsymbol{\pi}}{\mathrm{4}}−\boldsymbol{{cosx}}.\boldsymbol{{sin}}\frac{\boldsymbol{\pi}}{\mathrm{4}}\right)+\boldsymbol{{c}} \\$$$$\:\:\:=\frac{\boldsymbol{{e}}^{\boldsymbol{{x}}} }{\sqrt{\mathrm{2}}}\boldsymbol{{sin}}\left(\boldsymbol{{x}}−\frac{\boldsymbol{\pi}}{\mathrm{4}}\right)+\boldsymbol{{c}}\:\:\:\:\:\:\:\:\boldsymbol{{ans}} \\$$$$\boldsymbol{{or}}\:=\frac{\boldsymbol{{e}}^{\boldsymbol{{x}}} }{\sqrt{\mathrm{2}}}\boldsymbol{{cos}}\left[\frac{\boldsymbol{\pi}}{\mathrm{2}}+\left(\boldsymbol{{x}}−\frac{\boldsymbol{\pi}}{\mathrm{4}}\right)\right]+\boldsymbol{{c}} \\$$$$\:\:\:\:=\frac{\boldsymbol{{e}}^{\boldsymbol{{x}}} }{\sqrt{\mathrm{2}}}\boldsymbol{{cos}}\left[\boldsymbol{{x}}+\frac{\boldsymbol{\pi}}{\mathrm{4}}\right]+\boldsymbol{{c}}\:\:\:\:\:\:\:\:\boldsymbol{{ans}} \\$$$$\boldsymbol{{let}}\:\:\: \\$$$$\boldsymbol{{I}}=\int\boldsymbol{{e}}^{\boldsymbol{{x}}} \boldsymbol{{sinx}}=\boldsymbol{{e}}^{\boldsymbol{{x}}} \boldsymbol{{sinx}}−\int\boldsymbol{{e}}^{\boldsymbol{{x}}} \boldsymbol{{cosx}} \\$$$$\:\:\:\:\:\:\:\:\:\:\:=\boldsymbol{{e}}^{\boldsymbol{{x}}} \boldsymbol{{sinx}}−\boldsymbol{{e}}^{\boldsymbol{{x}}} \boldsymbol{{cosx}}−\int\boldsymbol{{e}}^{\boldsymbol{{x}}} \boldsymbol{{sinx}} \\$$$$\:\:\:\:\:\:\:\:\:\:\:=\boldsymbol{{e}}^{\boldsymbol{{x}}} \left(\boldsymbol{{sinx}}−\boldsymbol{{cosx}}\right)−\boldsymbol{{I}}+\boldsymbol{{k}} \\$$$$\mathrm{2}\boldsymbol{{I}}=\boldsymbol{{e}}^{\boldsymbol{{x}}} \left(\boldsymbol{{sinx}}−\boldsymbol{{cosx}}\right)+\boldsymbol{{k}} \\$$$$\boldsymbol{{so}}\:\boldsymbol{{I}}=\frac{\boldsymbol{{e}}^{\boldsymbol{{x}}} }{\mathrm{2}}\left(\boldsymbol{{sinx}}−\boldsymbol{{cosx}}\right)+\boldsymbol{{c}}\:\:\:\:\left[\boldsymbol{{c}}=\frac{\boldsymbol{{k}}}{\mathrm{2}}=\boldsymbol{{constant}}\right] \\$$

Commented by smridha last updated on 01/Jul/20

$$\boldsymbol{{who}}\:\boldsymbol{{are}}\:\boldsymbol{{you}}\:\boldsymbol{{man}}\:??\boldsymbol{{well}}\:\boldsymbol{{copied}}!! \\$$

Answered by MJS last updated on 01/Jul/20

$$\mathrm{why}\:\mathrm{so}\:\mathrm{complicated}? \\$$$$\mathrm{it}'\mathrm{s}\:\mathrm{simply}\:\mathrm{2}\:\mathrm{times}\:\mathrm{by}\:\mathrm{parts} \\$$$${I}=\int\mathrm{e}^{{x}} \mathrm{sin}\:{x}\:{dx}= \\$$$$\:\:\:\:\:{u}'={u}=\mathrm{e}^{{x}} \\$$$$\:\:\:\:\:{v}=\mathrm{sin}\:{x}\:\rightarrow\:{v}'=\mathrm{cos}\:{x} \\$$$$=\mathrm{e}^{{x}} \mathrm{sin}\:{x}\:−\int\mathrm{e}^{{x}} \mathrm{cos}\:{x}\:{dx}= \\$$$$\:\:\:\:\:{u}'={u}=\mathrm{e}^{{x}} \\$$$$\:\:\:\:\:{v}=\mathrm{cos}\:{x}\:\rightarrow\:{v}'=−\mathrm{sin}\:{x} \\$$$$=\mathrm{e}^{{x}} \mathrm{sin}\:{x}\:−\left(\mathrm{e}^{{x}} \mathrm{cos}\:{x}\:−\int−\mathrm{e}^{{x}} \mathrm{sin}\:{x}\:{dx}\right)= \\$$$$=\mathrm{e}^{{x}} \mathrm{sin}\:{x}\:−\mathrm{e}^{{x}} \mathrm{cos}\:{x}\:−\int\mathrm{e}^{{x}} \mathrm{sin}\:{x}\:{dx} \\$$$$\mathrm{now}\:\mathrm{we}\:\mathrm{have} \\$$$${I}=\mathrm{e}^{{x}} \mathrm{sin}\:{x}\:−\mathrm{e}^{{x}} \mathrm{cos}\:{x}\:−{I} \\$$$$\mathrm{2}{I}=\mathrm{e}^{{x}} \mathrm{sin}\:{x}\:−\mathrm{e}^{{x}} \mathrm{cos}\:{x} \\$$$${I}=\frac{\mathrm{e}^{{x}} \mathrm{sin}\:{x}\:−\mathrm{e}^{{x}} \mathrm{cos}\:{x}}{\mathrm{2}}+{C} \\$$

Commented by smridha last updated on 02/Jul/20

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