int cos(ln x) dx
Consider instead:
int cos(ln x ) + i sin(ln x ) dx
Chosen because we can use of Euler's Formula
= int e^(i ln x ) dx
= int e^( ln x^i ) dx
= int x^i dx
= x^(i + 1)/(i + 1) + C
And the we reverse back into the original form:
= e^( (ln x^(i+1)) )/(i+1) + C
= e^((i + 1) (ln x ) )/(i+1) + C
= (e^(i ln(x)) e^(ln(x)) )/(i+1) + C
= (( cos(ln x ) + i sin(ln x )) x )/(i+1) + C
Use the conjugate of the denominator:
= ((1-i)( cos(ln x) + i sin(ln x )) x )/((i+1)(- i + 1)) + C
= (( cos(ln x) + i sin(ln x ) - i cos (ln x) + sin (ln x)) x )/2 + C
= (x ( cos(ln x) + sin (ln x)) )/2 + (i x( sin(ln x) - cos(ln x ) ))/2 + C
We started with a real integrand so we take the first term and the constant:
implies int cos(ln x) dx = (x ( cos(ln x) + sin (ln x)) )/2 + C
