Find the integral of:
csc(x) * dx
csc(x) = cos(x) / (cos(x)^2)
u = sin(x)
du = cos(x)dx
now using the trigonometric identity sin^2(x) + cos^2(x) = 1
we arrive at the following equivalent integral
du/(1-u^2)
which can be written as:
0.5*[1/(1-u) + 1/(1+u)]du
integrating we get:
0.5*[ln(1+u) - ln(1-u)] = 0.5*ln[(1+u)/(1-u)]
My first method is a bit long, but I propose double substitution of:
, so
You will then have
To solve for cosine in terms of u, use the sine, cosine relationship:
You will then have an integrand of
To simplify that, you need to make the substitution:
The resulting integrand becomes:
Integrating that, you get:
Substitution of leads to...
Substitution of leads to...
Hmmm, doesn't look like the answer I should get?
sorry my mistake, but the solution changes only slightly
csc(x) = sin(x) / (sin(x)^2)
u = cos(x)
du = -sin(x)dx
now using the trigonometric identity sin^2(x) + cos^2(x) = 1
we arrive at the following equivalent integral
-du/(1-u^2)
which can be written as:
-0.5*[1/(1-u) + 1/(1+u)]du
integrating we get:
-0.5*[ln(1+u) - ln(1-u)] = 0.5*ln[(1-u)/(1+u)] + const =
0.5*ln[(1-cos(x))/(1+cos(x))] + const