Definite integral tg³x from 0 to π/3

This problem asks to evaluate the definite integral of the cube of the tangent on the interval [0, π/3].

Problem

$\int_{0}^{π / 3} {tg}^{3} x d x$

Solution

Step 1

Express tg³x in terms of sin and cos:

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

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

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

Step 2

Rewrite the integral:

How do we bring sin x under the differential?

We know that d(cos x) = −sin x dx. This means sin x dx = −d(cos x). By isolating one factor sin x in the numerator, we can replace sin x dx with −d(cos x). Then the remaining sin²x and cos³x are expressed in terms of cos x (using the basic trigonometric identity), and the integral is evaluated as a power of cos x.

\int_{0}^{π / 3} \frac{{sin}^{3} x}{{cos}^{3} x} d x

Step 3

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

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

\sin x d x = - d (cos x)

Step 4

After substitution we get:

- \int_{0}^{π / 3} \frac{1 - {cos}^{2} x}{{cos}^{3} x} d (cos x)

Step 5

Split into two integrals:

\int_{0}^{π / 3} (\frac{1}{cos x} - \frac{1}{{cos}^{3} x}) d (cos x)

Step 6

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

[_{0}^{π / 3} ln | cos x | + \frac{1}{2 {cos}^{2} x}

Step 7

Substitute the limits. cos 0 = 1, cos(π/3) = 1/2. We obtain the answer.

Answer

$\frac{3}{2} - ln 2$

Frequently asked questions

How do I evaluate a definite integral of tg³x?

Express tg³x = sin³x/cos³x. Isolate one factor sin x and bring it under the differential: sin x dx = −d(cos x). Replace the remaining sin²x by 1 − cos²x (basic trigonometric identity). Then the integral reduces to a power of cos 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 = cos x, so ln|cos x| appears.

What about odd powers of the tangent in general?

The standard technique: tg^(2n+1)x = tg^(2n−1)x · tg²x = tg^(2n−1)x · (1/cos²x − 1). Alternatively: tg^(2n+1)x = sin^(2n+1)x / cos^(2n+1)x. Isolate one sin x and bring it under the differential, as in this problem. Then express the remaining sin^(2n)x in terms of cos²x using the basic trigonometric identity.

Why do the limits of integration not change under the substitution?

We did not make an explicit change of variable — we simply brought sin x dx under the differential as −d(cos x). The limits in x remain the same: from 0 to π/3. When we substitute them into the expression in terms of cos x, we compute cos 0 = 1 and cos(π/3) = 1/2.