Definite integral ctg³x from π/6 to π/4

This problem asks to evaluate the definite integral of the cube of the cotangent on the interval [π/6, π/4].

Problem

$\int_{π / 6}^{π / 4} {ctg}^{3} x d x$

Solution

Step 1

Express ctg³x in terms of sin and cos:

Why do we express the cotangent in terms of sine and cosine?

For odd powers of ctg x (or tg x), the standard technique is to express the cotangent as cos x/sin x. This allows us to isolate the factor cos x, which can be brought under the differential (since d(sin x) = cos x dx). The subsequent substitution t = sin x reduces the integral to a tabular form.

{ctg}^{3} x = \frac{{cos}^{3} x}{{sin}^{3} x}

Step 2

Rewrite the integral:

How do we bring cos x under the differential?

We know that d(sin x) = cos x dx. This means cos x dx = d(sin x). By isolating one factor cos x in the numerator, we can replace cos x dx with d(sin x). Then the remaining cos²x and sin³x are expressed in terms of sin x (using the basic trigonometric identity), and the integral is evaluated as a power of sin x.

\int_{π / 6}^{π / 4} \frac{{cos}^{3} x}{{sin}^{3} x} d x

Step 3

Apply the basic trigonometric identity cos²x = 1 − sin²x and note that cos x dx = d(sin x):

{cos}^{2} x = 1 - {sin}^{2} x

cos x d x = d (sin x)

Step 4

After substitution we get:

\int_{π / 6}^{π / 4} \frac{1 - {sin}^{2} x}{{sin}^{3} x} d (sin x)

Step 5

Integrate: ∫ u⁻³ du = −1/(2u²), ∫ du/u = ln|u|. The antiderivative is −1/(2sin²x) − ln|sin x|. Apply the Newton-Leibniz formula:

[_{π / 6}^{π / 4} - \frac{1}{2 {sin}^{2} x} - ln | sin x |

Step 6

Substitute the limits. sin(π/4) = √2/2, sin(π/6) = 1/2. After simplification we obtain the answer.

Answer

$1 - \frac{1}{2} ln 2$

Frequently asked questions

How do I evaluate a definite integral of ctg³x?

Express ctg³x = cos³x/sin³x. Isolate one factor cos x and bring it under the differential: cos x dx = d(sin x). Replace the remaining cos²x by 1 − sin²x (basic trigonometric identity). Then the integral reduces to a power of sin x and is evaluated elementarily.

Why does the answer contain a logarithm?

Because when splitting into two integrals, one of the terms has the form ∫ du/u, and this integral is ln|u| + C. In our case u = sin x, so ln|sin x| appears.

Why can the integral not be taken from 0 to π/4?

As x → 0+, the function ctg³x → +∞, since ctg x = cos x/sin x and sin 0 = 0. This means the integrand has a singularity at x = 0, and the integral would become improper. In our problem the lower limit is π/6, where sin(π/6) = 1/2 ≠ 0, so the integral is proper.

What about odd powers of the cotangent in general?

The standard technique: ctg^(2n+1)x = cos^(2n+1)x / sin^(2n+1)x. Isolate one cos x and bring it under the differential, as in this problem. Then express the remaining cos^(2n)x in terms of sin²x using the basic trigonometric identity.