Question Number 221541 by PaulDirac last updated on 07/Jun/25
$$\int_{\mathrm{0}} ^{\:\pi} \mathrm{tan}\left(\mathrm{4}!\right)^{{e}^{\mathrm{3}{x}^{\mathrm{2}} +\:\mathrm{2}{x}} } .{dx} \\ $$
Answered by MrGaster last updated on 07/Jun/25
![=∫_0 ^( π) tan(24)^e^(3x^2 + 2x) .dx =∫_0 ^∞ exp(ln(tan(24))∙e^(3x^2 +2x) )dx =Σ_(n=0) ^∞ (((ln(tan(24)))^n )/(n!))∫_0 ^π e^(n(3x^2 +2x)) dx ∫_0 ^π e^(n(3x^2 +2x)) dx { ((π n=0)),((e^(−(n/3) ) n≥1)) :} ∫_(1/3) ^(π+(1/3)) e^(3na^2 ) du=[((√π)/(2(√(3n))))erfi((√(3n))u)]_(1/3) ^(π+(1/3)) =((√π)/(2(√(3n))))(erfi((√(3n))(π+(1/3)))−erfi((√(3n))∙(1/3))) ⇒∫_0 ^π e^(n(3x^2 +2x)) dx=e^(−(n/3)) ∙((√π)/(2(√(3n))))(erfi((√(3n))(π+(1/3)))−erfi((√(3n))∙(1/3))) ⇒Σ_(n=0) ^∞ (((ln(tan(24)))^n )/(n!))∫_(0 ) ^π e^(n(3x^2 +2x)) dx=π+Σ_(n=1) ^∞ (((ln(tan(24)))^n e^(−(n/3)) (√π))/(n!2(√3)(√n)))(erfi((√(3n))(π+(1/3)))−erfi((√(3n))∙(1/n)) ∴∫_0 ^π tan(24)e^(3x^2 +2x) dx=π+((√π)/(2(√3)))Σ_(n=1) ^∞ (((ln(tan(24))∙e^(−(1/3)) )^n )/(n!n^(1/2) ))(erfi((√3)(π+(1/3))(√n))−erfi(((√n)/( (√3)))))](https://www.tinkutara.com/question/Q221545.png)
$$=\int_{\mathrm{0}} ^{\:\pi} \mathrm{tan}\left(\mathrm{24}\right)^{{e}^{\mathrm{3}{x}^{\mathrm{2}} +\:\mathrm{2}{x}} } .{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \mathrm{exp}\left(\mathrm{ln}\left(\mathrm{tan}\left(\mathrm{24}\right)\right)\centerdot{e}^{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}} \right){dx} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{ln}\left(\mathrm{tan}\left(\mathrm{24}\right)\right)\right)^{{n}} }{{n}!}\int_{\mathrm{0}} ^{\pi} {e}^{{n}\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}\right)} {dx} \\ $$$$\int_{\mathrm{0}} ^{\pi} {e}^{{n}\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}\right)} {dx\begin{cases}{\pi\:\:\:\:\:\:\:{n}=\mathrm{0}}\\{{e}^{−\frac{{n}}{\mathrm{3}}\:\:} {n}\geq\mathrm{1}}\end{cases}} \\ $$$$\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\pi+\frac{\mathrm{1}}{\mathrm{3}}} {e}^{\mathrm{3}{na}^{\mathrm{2}} } {du}=\left[\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{3}{n}}}\mathrm{erfi}\left(\sqrt{\mathrm{3}{n}}{u}\right)\right]_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\pi+\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{3}{n}}}\left(\mathrm{erfi}\left(\sqrt{\mathrm{3}{n}}\left(\pi+\frac{\mathrm{1}}{\mathrm{3}}\right)\right)−\mathrm{erfi}\left(\sqrt{\mathrm{3}{n}}\centerdot\frac{\mathrm{1}}{\mathrm{3}}\right)\right) \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\pi} {e}^{{n}\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}\right)} {dx}={e}^{−\frac{{n}}{\mathrm{3}}} \centerdot\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{3}{n}}}\left(\mathrm{erfi}\left(\sqrt{\mathrm{3}{n}}\left(\pi+\frac{\mathrm{1}}{\mathrm{3}}\right)\right)−\mathrm{erfi}\left(\sqrt{\mathrm{3}{n}}\centerdot\frac{\mathrm{1}}{\mathrm{3}}\right)\right) \\ $$$$\Rightarrow\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{ln}\left(\mathrm{tan}\left(\mathrm{24}\right)\right)\right)^{{n}} }{{n}!}\int_{\mathrm{0}\:} ^{\pi} {e}^{{n}\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}\right)} {dx}=\pi+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{ln}\left(\mathrm{tan}\left(\mathrm{24}\right)\right)\right)^{{n}} {e}^{−\frac{{n}}{\mathrm{3}}} \sqrt{\pi}}{{n}!\mathrm{2}\sqrt{\mathrm{3}}\sqrt{{n}}}\left(\mathrm{erfi}\left(\sqrt{\mathrm{3}{n}}\left(\pi+\frac{\mathrm{1}}{\mathrm{3}}\right)\right)−\mathrm{erfi}\left(\sqrt{\mathrm{3}{n}}\centerdot\frac{\mathrm{1}}{{n}}\right)\right. \\ $$$$\therefore\int_{\mathrm{0}} ^{\pi} \mathrm{tan}\left(\mathrm{24}\right){e}^{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}} {dx}=\pi+\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{3}}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{ln}\left(\mathrm{tan}\left(\mathrm{24}\right)\right)\centerdot{e}^{−\frac{\mathrm{1}}{\mathrm{3}}} \right)^{{n}} }{{n}!{n}^{\frac{\mathrm{1}}{\mathrm{2}}} }\left(\mathrm{erfi}\left(\sqrt{\mathrm{3}}\left(\pi+\frac{\mathrm{1}}{\mathrm{3}}\right)\sqrt{{n}}\right)−\mathrm{erfi}\left(\frac{\sqrt{{n}}}{\:\sqrt{\mathrm{3}}}\right)\right) \\ $$