#int (1+sinx)/(sinx * (1+cosx))dx# ?

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Answer 1 · 2018-04-03T18:09:15

#1/4(tan(x/2))^2+tan(x/2)+1/2Ln(tan(x/2))+C#

#int (1+sinx)/(sinx*(1+cosx))*dx#

After using #y=tan(x/2)#, #dx=(2dy)/(y^2+1)#, #sinx=(2y)/(y^2+1)# and #cosx=(1-y^2)/(y^2+1)# transforms, this integral became

#int (1+(2y)/(y^2+1))/((2y)/(y^2+1)(1+(1-y^2)/(y^2+1)))*(2dy)/(y^2+1)#

=#int ((y^2+2y+1)/(y^2+1))/((2y)/(y^2+1)2/(y^2+1))*(2dy)/(y^2+1)#

=#int ((y^2+2y+1)*dy)/(2y)#

=#1/2int y*dy+int dy+1/2int (dy)/y#

=#y^2/4+y+1/2Lny+C#

=#1/4(tan(x/2))^2+tan(x/2)+1/2Ln(tan(x/2))+C#