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Question Number 197436 by horsebrand11 last updated on 17/Sep/23

∫_0 ^(π/2) (dx/(3+tan x)) =?

$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{\mathrm{dx}}{\mathrm{3}+\mathrm{tan}\:\mathrm{x}}\:=? \\ $$

Answered by Frix last updated on 17/Sep/23

∫(dx/(3+tan x)) =^(t=tan x) ∫(dt/((t+3)(t^2 +1)))= =(1/(10))∫((1/(t+3))+(3/(t^2 +1))−(t/(t^2 +1)))dt= =(1/(10))(ln (t+3) +3tan^(−1) t −(1/2)ln (t^2 +1))= =((3x)/(10))+((ln ∣3cos x +sin x∣)/(10))+C ⇒ Answer is ((3π)/(20))−((ln 3)/(10))

$$\int\frac{{dx}}{\mathrm{3}+\mathrm{tan}\:{x}}\:\overset{{t}=\mathrm{tan}\:{x}} {=}\:\int\frac{{dt}}{\left({t}+\mathrm{3}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}\int\left(\frac{\mathrm{1}}{{t}+\mathrm{3}}+\frac{\mathrm{3}}{{t}^{\mathrm{2}} +\mathrm{1}}−\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\right){dt}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{10}}\left(\mathrm{ln}\:\left({t}+\mathrm{3}\right)\:+\mathrm{3tan}^{−\mathrm{1}} \:{t}\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({t}^{\mathrm{2}} +\mathrm{1}\right)\right)= \\ $$$$=\frac{\mathrm{3}{x}}{\mathrm{10}}+\frac{\mathrm{ln}\:\mid\mathrm{3cos}\:{x}\:+\mathrm{sin}\:{x}\mid}{\mathrm{10}}+{C} \\ $$$$\Rightarrow\:\mathrm{Answer}\:\mathrm{is}\:\frac{\mathrm{3}\pi}{\mathrm{20}}−\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{10}} \\ $$

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