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Question-190325




Question Number 190325 by Shrinava last updated on 31/Mar/23
Answered by mehdee42 last updated on 31/Mar/23
I=∫_0 ^(1/2)  ((2tan(β/2)x)/(αtan^2 (β/2)x+2tan(β/2)x+α))dx  if  tan(β/2)x=u⇒I=∫_0 ^(tan(β/4))  ((4du)/(αu^2 +2u+α))  it can be solved
$${I}=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{\mathrm{2}{tan}\frac{\beta}{\mathrm{2}}{x}}{\alpha{tan}^{\mathrm{2}} \frac{\beta}{\mathrm{2}}{x}+\mathrm{2}{tan}\frac{\beta}{\mathrm{2}}{x}+\alpha}{dx} \\ $$$${if}\:\:{tan}\frac{\beta}{\mathrm{2}}{x}={u}\Rightarrow{I}=\int_{\mathrm{0}} ^{{tan}\frac{\beta}{\mathrm{4}}} \:\frac{\mathrm{4}{du}}{\alpha{u}^{\mathrm{2}} +\mathrm{2}{u}+\alpha} \\ $$$${it}\:{can}\:{be}\:{solved} \\ $$

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