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:
$\displaystyle u=sin(x)$, so $\displaystyle csc(x) = \frac{1}{u}$
You will then have $\displaystyle du=cos(x)dx$
To solve for cosine in terms of u, use the sine, cosine relationship:
$\displaystyle cos^2(x)+sin^2(x)=1$
$\displaystyle cos^2(x)=1-u^2$
$\displaystyle cos(x)=\sqrt{1-u^2}$
You will then have an integrand of $\displaystyle \frac{1}{u\sqrt{1-u^2}}$
To simplify that, you need to make the substitution: $\displaystyle v=1-u^2$
$\displaystyle dv=-2udu$
$\displaystyle u=\sqrt{v-1}$
The resulting integrand becomes:
$\displaystyle \frac{-2}{\sqrt{v}} = -2v^{-1/2}$
Integrating that, you get: $\displaystyle -4v^{1/2}$
Substitution of $\displaystyle v=1-u^2$ leads to...
$\displaystyle -4{(1-u^2)}^{1/2}$
Substitution of $\displaystyle u=sin(x)$ leads to...
$\displaystyle -4{(1-sin^2(x))}^{1/2} = -4\sqrt{cos(x)}$
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