Question Number 149081 by Integrals last updated on 02/Aug/21

Answered by Ar Brandon last updated on 02/Aug/21

φ=∫(dβ/(2−3sinβ)) , t=tan(β/2) =∫(2/(2−3(((2t)/(1+t^2 )))))∙(dt/(1+t^2 ))=∫(dt/(t^2 −3t+1)) =∫(dt/((t−(3/2))^2 −(5/4)))=−(2/( (√5)))arctanh(((2t−3)/( (√5))))+C =(1/( (√5)))ln∣((2t−3−(√5))/(2t−3+(√5)))∣+C=((√5)/5)ln∣((2tan((β/2))−3−(√5))/(2tan((β/2))−3+(√5)))∣+C

Commented byAr Brandon last updated on 02/Aug/21

sinβ=((sinβ)/1)=((2sin(β/2)cos(β/2))/(cos^2 (β/2)+sin^2 (β/2)))=((2tan(β/2))/(1+tan^2 (β/2))) t=tan(β/2)⇒dt=(1/2)sec^2 (β/2)dβ=(1/2)(1+tan^2 (β/2))dβ arctanh(x)=(1/2)ln∣((1+x)/(1−x))∣