Question

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)`

Answer 1

`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`