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
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