Integration Questions

Question Number 89382 by M±th+et£s last updated on 17/Apr/20

$$\int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{2}{x}} \:{cos}\left(\mathrm{3}{x}\right)\:{sin}\left(\mathrm{4}{x}\right) \\$$

Commented by mathmax by abdo last updated on 17/Apr/20

$${cos}\left(\mathrm{3}{x}\right){sin}\left(\mathrm{4}{x}\right)\:={cos}\left(\mathrm{3}{x}\right){cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{4}{x}\right) \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\mathrm{3}{x}+\frac{\pi}{\mathrm{2}}−\mathrm{4}{x}\right)\:+{cos}\left(\mathrm{3}{x}−\frac{\pi}{\mathrm{2}}+\mathrm{4}{x}\right)\right\} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\frac{\pi}{\mathrm{2}}−{x}\right)\:+{cos}\left(\mathrm{7}{x}−\frac{\pi}{\mathrm{2}}\right)\right\} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{sinx}\:+{sin}\left(\mathrm{7}{x}\right)\right\}\:\Rightarrow \\$$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−\mathrm{2}{x}} \:{cos}\left(\mathrm{3}{x}\right){sin}\left(\mathrm{4}{x}\right){dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{2}{x}} \:{sinx}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{2}{x}} \:{sin}\left(\mathrm{7}{x}\right){dx} \\$$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−\mathrm{2}{x}} \:{sinx}\:{dx}\:={Im}\left(\int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{2}{x}\:+{ix}} {dx}\right) \\$$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{\left(−\mathrm{2}+{i}\right){x}} \:{dx}\:=\left[\frac{\mathrm{1}}{−\mathrm{2}+{i}}{e}^{\left(−\mathrm{2}+{i}\right){x}} \right]_{\mathrm{0}} ^{+\infty} \:=\frac{\mathrm{1}}{−\mathrm{2}+{i}}\left(−\mathrm{1}\right) \\$$$$=\frac{\mathrm{1}}{\mathrm{2}−{i}}\:=\frac{\mathrm{2}+{i}}{\mathrm{5}}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:{e}^{−\mathrm{2}{x}} \:{sinxdx}\:=\frac{\mathrm{1}}{\mathrm{5}} \\$$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−\mathrm{2}{x}} {sin}\left(\mathrm{7}{x}\right){dx}\:={Im}\left(\int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{2}{x}+\mathrm{7}{ix}} {dx}\right) \\$$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{\left(−\mathrm{2}+\mathrm{7}{i}\right){x}} \:{dx}\:=\left[\frac{\mathrm{1}}{−\mathrm{2}+\mathrm{7}{i}}{e}^{\left(−\mathrm{2}+\mathrm{7}{i}\right){x}} \right]_{\mathrm{0}} ^{+\infty} \:=\frac{\mathrm{1}}{−\mathrm{2}+\mathrm{7}{i}}\left(−\mathrm{1}\right) \\$$$$=\frac{\mathrm{1}}{\mathrm{2}−\mathrm{7}{i}}\:=\frac{\mathrm{2}+\mathrm{7}{i}}{\mathrm{4}+\mathrm{49}}\:=\frac{\mathrm{2}+\mathrm{7}{i}}{\mathrm{53}}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{2}{x}} {sin}\left(\mathrm{7}{x}\right){dx}=\frac{\mathrm{7}}{\mathrm{53}}\:\Rightarrow \\$$$${I}\:=\frac{\mathrm{1}}{\mathrm{10}}\:+\frac{\mathrm{7}}{\mathrm{106}}\:=\frac{\mathrm{106}+\mathrm{70}}{\mathrm{1060}}\:=\frac{\mathrm{176}}{\mathrm{1060}}\:=... \\$$

Commented by mathmax by abdo last updated on 18/Apr/20

$${cos}\left(\mathrm{3}{x}\right){sin}\left(\mathrm{4}{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({sinx}\:+{sin}\left(\mathrm{7}{x}\right)\:{not}\:\frac{\mathrm{1}}{\mathrm{2}}\left({sin}\left(\mathrm{7}{x}\right)−{sinx}\right)..!\right. \\$$

Answered by TANMAY PANACEA. last updated on 17/Apr/20

$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{2}{x}} \left({sin}\mathrm{7}{x}−{sinx}\right){dx} \\$$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{2}{x}} {sin}\mathrm{7}{xdx}−\int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{2}{x}} {sinx}\:{dx} \\$$$$\mathrm{2}{I}_{\bigstar} =\int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{2}{x}} {cos}\mathrm{7}{x}−\int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{2}{x}} {cosx}\:{dx} \\$$$$\mathrm{2}\left({I}_{\bigstar} +{iI}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{2}{x}} .{e}^{{i}\mathrm{7}{x}} {dx}−\int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{2}{x}} .{e}^{{ix}} {dx} \\$$$$\mathrm{2}\left({I}_{\bigstar} +{iI}\right)=\mid\frac{{e}^{−\mathrm{2}{x}+{i}\mathrm{7}{x}} }{−\mathrm{2}+{i}\mathrm{7}}−\frac{{e}^{−\mathrm{2}{x}+{ix}} }{−\mathrm{2}+{i}}\mid_{\mathrm{0}} ^{\infty} \\$$$$\mathrm{2}\left({I}_{\bigstar} +{I}\right)=\frac{\mathrm{1}}{−\mathrm{2}+{i}\mathrm{7}}−\frac{\mathrm{1}}{−\mathrm{2}+{i}}=\frac{−\mathrm{2}−{i}\mathrm{7}}{\mathrm{4}−\mathrm{49}}−\frac{−\mathrm{2}−{i}}{\mathrm{4}−\mathrm{1}} \\$$$$\mathrm{2}\left({I}_{\bigstar} +{I}\right)=\frac{−\mathrm{2}}{−\mathrm{45}}+\frac{−{i}\mathrm{7}}{−\mathrm{45}}+\frac{\mathrm{2}+{i}}{\mathrm{3}} \\$$$$\mathrm{2}\left({I}_{\bigstar} +{I}\right)=\frac{\mathrm{2}}{\mathrm{45}}+\frac{\mathrm{2}}{\mathrm{3}}+{i}\left(\frac{\mathrm{7}}{\mathrm{45}}+\frac{\mathrm{1}}{\mathrm{3}}\right) \\$$$$\mathrm{2}\left({I}_{\bigstar} +{I}\right)=\frac{\mathrm{2}+\mathrm{30}}{\mathrm{45}}+{i}\left(\frac{\mathrm{7}+\mathrm{15}}{\mathrm{45}}\right) \\$$$${I}_{\bigstar} +{I}=\frac{\mathrm{16}}{\mathrm{45}}+{i}×\frac{\mathrm{11}}{\mathrm{45}} \\$$$${I}=\frac{\mathrm{11}}{\mathrm{45}} \\$$$${pls}\:{check}\:{is}\:{it}\:{correct} \\$$

Commented by M±th+et£s last updated on 17/Apr/20

$${its}\:{correct}\:{sir}\:{thank}\:{you}\:{verry}\:{much} \\$$