How do you integrate? int [1/(sen^2x+4senx-5)]dx

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Answer 1 · 2018-03-12T21:14:58

#1/(3tan(x/2)-3)-sqrt6/36arctan((5tan(x/2)+1)/(2sqrt6))+C#

#int (dx)/[(sinx)^2+4sinx-5]#

=#-int (dx)/[5-4sinx-(sinx)^2]#

=#-1/6int (6dx)/[(5+sinx)*(1-sinx)]#

=#-1/6int dx/(5+sinx)#-#1/6int dx/(1-sinx)#

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

#-1/6int [(2dy)/(y^2+1)]/(5+(2y)/(y^2+1))#-#1/6int [(2dy)/(y^2+1)]/(1-(2y)/(y^2+1))#

=#-1/3int dy/(5y^2+2y+5)#-#1/3int dy/(y^2-2y+1)#

=#-1/3int (5dy)/(25y^2+10y+25)#-#1/3int dy/(y-1)^2#

=#1/3*(y-1)^(-1)#-#1/3int (5dy)/((5y+1)^2+24)#

=#1/(3y-3)-sqrt6/36arctan((5y+1)/(2sqrt6))+C#

=#1/(3tan(x/2)-3)-sqrt6/36arctan((5tan(x/2)+1)/(2sqrt6))+C#