Question Number 88859 by 242242864 last updated on 13/Apr/20
$$\int\:\boldsymbol{{e}}^{\boldsymbol{{ax}}} \mathrm{cos}\:\boldsymbol{{bx}}\:\boldsymbol{{dx}} \\ $$$$\int\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{e}}^{\mathrm{2}\boldsymbol{{x}}} \boldsymbol{{ln}}\mathrm{3}\boldsymbol{{x}}^{\mathrm{2}} \:\boldsymbol{{dx}} \\ $$
Commented by john santu last updated on 13/Apr/20
![1) I= ∫ e^(ax) cos bx dx I = (1/b)e^(ax) sin bx − (1/(ab))∫ e^(ax) sin bx dx I = ((e^(ax) sin bx)/b) − (1/(ab)) [ −((e^(ax) cos bx)/b) + (1/(ab))∫ e^(ax) cos bx dx ] I = ((e^(ax) sin bx)/b) + ((e^(ax) cos bx)/(ab^2 )) −(1/((ab)^2 )) ∫ e^(ax) cos bx dx (1+(1/((ab)^2 ))) I = ((e^(ax) (ab sin bx + cos bx))/(ab^2 )) I = ((a^2 b^2 e^(ax) (ab sin bx + cos bx ))/((1+(ab)^2 )(ab^2 ))) I= ((ae^(ax) (ab sin bx + cos bx ))/((1+(ab)^2 )))](Q88862.png)
$$\left.\mathrm{1}\right)\:{I}=\:\int\:{e}^{{ax}} \:\mathrm{cos}\:{bx}\:{dx}\: \\ $$$${I}\:=\:\frac{\mathrm{1}}{{b}}{e}^{{ax}} \:\mathrm{sin}\:{bx}\:−\:\frac{\mathrm{1}}{{ab}}\int\:{e}^{{ax}} \:\mathrm{sin}\:{bx}\:{dx} \\ $$$${I}\:=\:\frac{{e}^{{ax}} \:\mathrm{sin}\:{bx}}{{b}}\:−\:\frac{\mathrm{1}}{{ab}}\:\left[\:−\frac{{e}^{{ax}} \:\mathrm{cos}\:{bx}}{{b}}\:+\:\frac{\mathrm{1}}{{ab}}\int\:{e}^{{ax}} \:\mathrm{cos}\:{bx}\:{dx}\:\right] \\ $$$${I}\:=\:\frac{{e}^{{ax}} \:\mathrm{sin}\:{bx}}{{b}}\:+\:\frac{{e}^{{ax}} \:\mathrm{cos}\:{bx}}{{ab}^{\mathrm{2}} }\:−\frac{\mathrm{1}}{\left({ab}\right)^{\mathrm{2}} }\:\int\:{e}^{{ax}} \:\mathrm{cos}\:{bx}\:{dx}\: \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\left({ab}\right)^{\mathrm{2}} }\right)\:{I}\:=\:\frac{{e}^{{ax}} \:\left({ab}\:\mathrm{sin}\:{bx}\:+\:\mathrm{cos}\:{bx}\right)}{{ab}^{\mathrm{2}} } \\ $$$${I}\:=\:\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} \:{e}^{{ax}} \:\left({ab}\:\mathrm{sin}\:{bx}\:+\:\mathrm{cos}\:{bx}\:\right)}{\left(\mathrm{1}+\left({ab}\right)^{\mathrm{2}} \right)\left({ab}^{\mathrm{2}} \right)} \\ $$$${I}=\:\frac{{ae}^{{ax}} \:\left({ab}\:\mathrm{sin}\:{bx}\:+\:\mathrm{cos}\:{bx}\:\right)}{\left(\mathrm{1}+\left({ab}\right)^{\mathrm{2}} \right)}\: \\ $$