How can i find the integral of cot^3x in two different ways?

Answer 1

Integral (cot^3(x) ) dx

Here's one way: split up cot^3(x) into cot^2(x) and cot(x)

Integral ( (cot^2(x) cot(x)) dx )

cot^2(x) = csc^2(x) - 1, so we have

Integral ( (csc^2(x) - 1)cot(x) dx )

Distribute the cot(x) within the integral.

Integral ( csc^2(x)cot(x) - cot(x) ) dx

We can now split this up into two integrals.

Integral (cot(x) csc^2(x) dx) - Integral (cot(x)) dx

In the first integral, we *almost* have a product of a function and its integral, because the integral of cot(x) is -csc^2(x). We can offset this by adding a minus sign, which we in turn offset by putting a (-1) outside of the integral.

(-1) Integral (cot(x) (-csc^2(x)) dx ) - Integral (cot(x)) dx

Whenever we have a function next to its derivative, and we integrate, then Integral (f(x) f'(x) dx) is equal to (1/2)[f(x)]^2. Therefore, we have

(-1) (1/2) [cot^2(x)] - Integral (cot(x) dx)

But cot(x) = cos(x) / sin(x)

(-1) (1/2) [cot^2(x)] - Integral (cos(x)/sin(x))dx

We can do u substitution on the second integral; let u = sin(x). Then du = cos(x) dx.

(-1) (1/2) [cot^2(x)] - Integral (1/u)du

(-1) (1/2) [cot^2(x)] - ln|u| + C

But u = sin(x), so we have

(-1) (1/2) [cot^2(x)] - ln|sin(x)| + C

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Here's a second way

Integral (cot^3(x) dx)

Change to sines and cosines.

Integral ( [cos^3(x)] / [sin^3(x)] ) dx

Pull out a cosine from the numerator.

Integral ( [ cos^2(x) cos(x) ] / [sin^3(x)] ) dx

Now, using the identity cos^2(x) = 1 - sin^2(x)

Integral ( [ (1 - sin^2(x)) cos(x) / [sin^3(x)] ) dx

At this point, we use substitution.
Let u = sin(x). Then du = cos(x) dx, and we have

Integral ( (1 - u^2) / u^3 ) du

Splitting this up into two fractions, we have

Integral ( (1/u^3) - (u^2/u^3) ) du

Integral ( u^(-3) - 1/u ) du

Now we can integrate each of these directly.

u^(-2) / (-2) - ln|u| + C

Changing everything to positive exponents,

(1/u^2) (-1/2) - ln|u| + C

But u = sin(x), so

1/sin^2(x) (-1/2) - ln|sin(x)| + C