# Thread: integral (trig)

1. ## integral (trig)

Find the integral of:

csc(x) * dx

2. 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)]

3. Originally Posted by Peritus
csc(x) = cos(x) / (cos(x)^2)
I'd re-think that.

--

About the integral, multiply it by $\displaystyle \csc x-\cot x.$

4. Originally Posted by DINOCALC09
Find the integral of:

csc(x) * dx
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?

5. 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

6. The book has the answer as:

-ln|csc(x)+cot(x)| + c

The only problem is that it doesn't show me how to get there.

7. -ln[csc(x)+cot(x)]=-ln[(1/sin(x) + cos(x)/sin(x)] = ln[sin(x)/(1+cos(x))]

<=> 0.5*ln[sin^2(x) / ((1+cos(x))^2)] =
= 0.5*ln[(1-cos^2(x)) / ((1+cos(x))^2)] =
= 0.5*ln{[(1-cos(x))*(1+cos(x))] / ((1+cos(x))^2)} = 0.5*ln[(1-cos(x))/(1+cos(x))]