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)
Is it possible to solve this integrate using another way
not this way --> substituting u=
1 Answer
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
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
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
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