Question Number 89560 by peter frank last updated on 18/Apr/20
$${without}\:{use}\:{intergration} \\ $$$${by}\:{party} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {e}^{\theta} \mathrm{cos}\:\mathrm{2}\theta\:{d}\theta \\ $$
Commented by mathmax by abdo last updated on 18/Apr/20
![∫_0 ^(π/4) e^θ cos(2θ)dθ =Re(∫_0 ^(π/4) e^(θ+2iθ) dθ) and ∫_0 ^(π/4) e^((1+2i)θ) dθ =[(1/(1+2i))e^((1+2i)θ) ]_0 ^(π/4) =(1/(1+2i)){e^((1+2i)(π/4)) −1} =(((1−2i))/5)(e^(π/4) i−1) =(1/5)(e^(π/4) i−1+2e^(π/4) +2i) ⇒ ∫_0 ^(π/4) e^θ cos(2θ)dθ =((2e^(π/4) −1)/5)](Q89581.png)
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{e}^{\theta} \:{cos}\left(\mathrm{2}\theta\right){d}\theta\:={Re}\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{e}^{\theta+\mathrm{2}{i}\theta} \:{d}\theta\right)\:{and} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{e}^{\left(\mathrm{1}+\mathrm{2}{i}\right)\theta} \:{d}\theta\:=\left[\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}{i}}{e}^{\left(\mathrm{1}+\mathrm{2}{i}\right)\theta} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}{i}}\left\{{e}^{\left(\mathrm{1}+\mathrm{2}{i}\right)\frac{\pi}{\mathrm{4}}} −\mathrm{1}\right\} \\ $$$$=\frac{\left(\mathrm{1}−\mathrm{2}{i}\right)}{\mathrm{5}}\left({e}^{\frac{\pi}{\mathrm{4}}} \:{i}−\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{5}}\left({e}^{\frac{\pi}{\mathrm{4}}} {i}−\mathrm{1}+\mathrm{2}{e}^{\frac{\pi}{\mathrm{4}}} \:+\mathrm{2}{i}\right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{e}^{\theta} \:{cos}\left(\mathrm{2}\theta\right){d}\theta\:=\frac{\mathrm{2}{e}^{\frac{\pi}{\mathrm{4}}} −\mathrm{1}}{\mathrm{5}} \\ $$