How do you integrate this?

#int 1/(senx+5)# dx
Is it possible to solve this integrate using another way
not this way --> substituting u=#tan(x/2)#

1 Answer
Øko
Apr 5, 2018

#I=(arctan(cos(x)/(2sqrt(6)))+arctan((2sqrt(6)tan(x))/5))/(2sqrt(6))+C#

Explanation:

Yes, it's a little more complicated however, we seek

#I=int1/(sin(x)+5)dx#

Multiply the DEN and NUM by the conjugate #color(green)(sin(x)-5#

#I=int(sin(x)-5)/((sin(x)+5)(sin(x)-5))dx#

#color(white)(I)=int(sin(x)-5)/((sin^2(x)-25))dx#

#color(white)(I)=-int(sin(x))/((cos^2(x)+24))dx+int(5)/((cos^2(x)+24))dx#

First integral:

#I_1=-int(sin(x))/((cos^2(x)+24))dx#

Make a substitution #u=cos(x)=>du=-sin(x)dx#

#I_1=int1/((u^2+24))du#

#color(white)(I_1)=1/24int1/((u/sqrt(24))^2+1)du#

#color(white)(I_1)=arctan(cos(x)/sqrt(24))/sqrt(24)+C#

Second integral:

#I_2=int(5)/((cos^2(x)+24))dx#

#color(white)(I_2)=5int(sec^2(x))/((1+24sec^2(x)))dx#

#color(white)(I_2)=5int(sec^2(x))/(24tan^2(x)+25)dx#

Make a substitution #u=tan(x)=>du=sec^2(x)dx#

#I_2=5int1/(24u^2+25)du#

#color(white)(I_2)=1/5int1/(((sqrt(24)u)/5)^2+1)du#

#color(white)(I_2)=arctan((sqrt(24)tan(x))/5)/sqrt(24)+C#

Thus

#I=(arctan(cos(x)/sqrt(24))+arctan((sqrt(24)tan(x))/5))/sqrt(24)+C#

#color(white)(I)=(arctan(cos(x)/(2sqrt(6)))+arctan((2sqrt(6)tan(x))/5))/(2sqrt(6))+C#

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