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Question Number 66114 by Rio Michael last updated on 09/Aug/19

∫(((e^(2x) −sin2x)/(e^(2x) +cos2x)))dx = ?

$$\int\left(\frac{{e}^{\mathrm{2}{x}} −{sin}\mathrm{2}{x}}{{e}^{\mathrm{2}{x}} +{cos}\mathrm{2}{x}}\right){dx}\:=\:? \\ $$

Answered by $@ty@m123 last updated on 09/Aug/19

Let e^(2x) +cos 2x=z  ⇒ (2e^(2x) −2sin 2x)dx=dz  ∴ The given integral  ∫(dz/(2z))  =(1/2)ln z+C  =(1/2)ln ( e^(2x) +cos 2x)+C

$${Let}\:{e}^{\mathrm{2}{x}} +\mathrm{cos}\:\mathrm{2}{x}={z} \\ $$$$\Rightarrow\:\left(\mathrm{2}{e}^{\mathrm{2}{x}} −\mathrm{2sin}\:\mathrm{2}{x}\right){dx}={dz} \\ $$$$\therefore\:{The}\:{given}\:{integral} \\ $$$$\int\frac{{dz}}{\mathrm{2}{z}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:{z}+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\:{e}^{\mathrm{2}{x}} +\mathrm{cos}\:\mathrm{2}{x}\right)+{C} \\ $$$$ \\ $$

Commented by Rio Michael last updated on 09/Aug/19

i appreciate

$${i}\:{appreciate} \\ $$

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