Integration Technique: Substituting cot(u) to find the integral in terms of v

∫cot u du

To find the integral of cot(u) du, we can use the technique of integration by substitution

To find the integral of cot(u) du, we can use the technique of integration by substitution.

Let’s start by using the trigonometric identity: cot(u) = 1/tan(u).

Let’s make a substitution by letting v = tan(u), which implies du = dv / (1 + v^2). Now, we need to rewrite the entire integral in terms of v:

∫cot(u) du = ∫(1/tan(u)) du = ∫(1/v) (dv / (1 + v^2))

Now, we can split the integral into two parts:

∫(1/v) (dv / (1 + v^2)) = ∫(1 / (v(1 + v^2))) dv

To proceed, we can use a partial fraction decomposition. We can express 1 / (v(1 + v^2)) as:

1 / (v(1 + v^2)) = A/v + (Bv + C) / (1 + v^2)

Where A, B, and C are constants that we need to find. To find these constants, we can make the two expressions on the right-hand side have the same denominator:

1 = A(1 + v^2) + (Bv + C)v

Expanding and equating the coefficients of v^2, v, and the constant term, we get:

0v^2: 1 = A
1v^1: 0 = B
0v^0: 0 = A + C

From A = 1, we can find C = -A = -1.

Therefore, our partial fraction decomposition becomes:

1 / (v(1 + v^2)) = 1/v – (v + 1) / (1 + v^2)

Now, we can rewrite our integral using the partial fraction decomposition:

∫(1 / (v(1 + v^2))) dv = ∫(1/v – (v + 1) / (1 + v^2)) dv

Integrating each term separately:

= ∫(1/v) dv – ∫((v + 1) / (1 + v^2)) dv

The first integral ∫(1/v) dv is a standard integral and gives us ln|v|:

= ln|v| – ∫((v + 1) / (1 + v^2)) dv

The second integral can be evaluated using another substitution. Let’s substitute w = 1 + v^2, which implies dv = (dw) / (2v). Now, we can rewrite the integral in terms of w:

= ln|v| – ∫((v + 1) / (1 + v^2)) dv
= ln|v| – ∫(((w – 1) / 2v) / w) dw
= ln|v| – (1/2) ∫((w – 1) / (vw)) dw
= ln|v| – (1/2) ∫((w/v) – (1/v)) dw
= ln|v| – (1/2) ∫(1/v) dw + (1/2) ∫(1/w) dw
= ln|v| – (1/2) ln|v| + (1/2) ln|w| + C

Finally, substituting back v = tan(u), we get:

= ln|tan(u)| – (1/2) ln|tan(u)| + (1/2) ln|w| + C
= (1/2) ln|tan(u)| + (1/2) ln|w| + C
= (1/2) ln|tan(u)| + (1/2) ln| (1 + v^2) | + C

Therefore, the integral of cot(u) du is:

∫cot(u) du = (1/2) ln|tan(u)| + (1/2) ln| (1 + v^2) | + C

Where C is the constant of integration.

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