Hello I don't believe this was asked before but I was wondering the integral of cosx/sinx (or cotx)I think it is
let u=cosx du=-sinx dx -1/sinxdu=dx and then i would sinply take the integral of u? or would a make u= cosx and then end up with 1/u and the intergral sould be ln(sinx)
The hole problem I had was
(integral) x+3/x^4+9x^2 dx some algebra
(integral)x+3/9x^2(9x^2[{x^2/9}+1]})dx
so tan^2(theta)=x^2/9 tan(theta)=x/3 3tan(theta)=x dx=3sec^2(theta) remember the idenity 1+tan^2(x)=sec^2(x) so substituting gives
(integral) [3tan(theta)+3/81tan^2(theta)sec^2(theta)] 3sec^2(theta) d(theta)
some more algebra (integral)18/81tan(theta) d(theta)
Pulled the 18/81 out of the int and get (integral) 1/tan(theta) or cot(theta) or cos(theta)/sine(theta)
I believe sin over cos is ln of sin(theta) and the going back into terms of x would give the final answer of
ln[(sin{x/x^4+9x^2})/9]+c
thanks in advance for any help