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Question Number 113656 by bobhans last updated on 14/Sep/20

  ∫ (((1+tan (((3x)/2)))^2 )/(1+sin 3x)) dx ?

$$\:\:\int\:\frac{\left(\mathrm{1}+\mathrm{tan}\:\left(\frac{\mathrm{3x}}{\mathrm{2}}\right)\right)^{\mathrm{2}} }{\mathrm{1}+\mathrm{sin}\:\mathrm{3x}}\:\mathrm{dx}\:? \\ $$

Answered by john santu last updated on 14/Sep/20

 setting tan (((3x)/2)) = s →sin 3x = ((2s)/(1+s^2 ))  1+sin 3x = (((1+s)^2 )/(1+s^2 ))   (3/2) sec^2 (((3x)/2)) dx = ds   I = (2/3)∫ (((1+s)^2 )/(((((1+s)^2 )/(1+s^2 ))))) (ds/((1+s^2 )))  I = (2/3)∫ ds = (2/3) s + c   I= (2/3) tan (((3x)/2)) + c

$$\:{setting}\:\mathrm{tan}\:\left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right)\:=\:{s}\:\rightarrow\mathrm{sin}\:\mathrm{3}{x}\:=\:\frac{\mathrm{2}{s}}{\mathrm{1}+{s}^{\mathrm{2}} } \\ $$$$\mathrm{1}+\mathrm{sin}\:\mathrm{3}{x}\:=\:\frac{\left(\mathrm{1}+{s}\right)^{\mathrm{2}} }{\mathrm{1}+{s}^{\mathrm{2}} }\: \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{sec}\:^{\mathrm{2}} \left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right)\:{dx}\:=\:{ds}\: \\ $$$${I}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\int\:\frac{\left(\mathrm{1}+{s}\right)^{\mathrm{2}} }{\left(\frac{\left(\mathrm{1}+{s}\right)^{\mathrm{2}} }{\mathrm{1}+{s}^{\mathrm{2}} }\right)}\:\frac{{ds}}{\left(\mathrm{1}+{s}^{\mathrm{2}} \right)} \\ $$$${I}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\int\:{ds}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:{s}\:+\:{c}\: \\ $$$${I}=\:\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{tan}\:\left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right)\:+\:{c}\: \\ $$

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