Thread: integral (trig)

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)]

Originally Posted by Peritus

csc(x) = cos(x) / (cos(x)^2)

I'd re-think that.

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About the integral, multiply it by

Originally Posted by DINOCALC09

Find the integral of:

csc(x) * dx

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

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.

-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))]