Question

\(\quad 12-1012:(\pi / 9) \int_{0}(\tan x+\tan 2 x+\tan 3 x+\tan x \cdot \tan 2 x \cdot \tan 3 x) d x\) is equal to....... (a) \((1 / 3) \log 2\) (b) \(\log ^{3} \sqrt{4}\) (c) \(3 \log 2\) (d) \(4 \log \sqrt{3}\)

Short answer

The answer cannot be determined from the given information.

Step-by-step solution

Break the integral into parts

Sum and product of integrals can be expressed as multiple integrals with same limits: \[12 - 1012\left(\frac{\pi}{9}\right)\int_{0}^{\pi/9}(\tan x+\tan 2x+\tan 3x+\tan x \cdot \tan 2x \cdot \tan 3x) dx\] \[= 12 - 1012\left(\frac{\pi}{9}\right)\left[\int_{0}^{\pi/9}\tan xdx+\int_{0}^{\pi/9}\tan2xdx+\int_{0}^{\pi/9}\tan3xdx+\int_{0}^{\pi/9}\tan x \cdot \tan 2x \cdot \tan 3x dx\right]\]

Evaluate the individual integrals

Using antiderivative formulas, we can evaluate each integral: \[\int_{0}^{\pi/9}\tan x dx = \log|\sec x| \Big|_0^{\pi/9}\] \[\int_{0}^{\pi/9}\tan2x dx = \frac{1}{2}\log|\sec 2x| \Big|_0^{\pi/9}\] \[\int_{0}^{\pi/9}\tan3x dx = \frac{1}{3}\log|\sec 3x| \Big|_0^{\pi/9}\] For the product term, use the triple-angle formula for the tangent: \[\tan 3x = 3\tan x -\tan^3 x\] Rearrange it as: \[\tan x\tan 2x\tan 3x = (\tan^3 x)\left(\frac{\tan 3x}{\tan x}\right) = \tan^3x (3-\tan^3x)\] Now evaluate this integral: \[\int_{0}^{\pi/9}\tan x \cdot \tan 2x \cdot \tan 3x dx = \int_{0}^{\pi/9}(\tan^3x(3-\tan^3x)) dx\] This last integral is complicated to evaluate directly, but notice that at \(x = \pi/9\), both \(\tan x\) and \(\tan 2x\) are equal to 1, so their product with any other term will still be 0. Therefore, the integral is zero: \[\int_{0}^{\pi/9}\tan x \cdot \tan 2x \cdot \tan 3x dx = 0\]

Substitute integral values into the original expression

Substitute the evaluated integrals: \[12 - 1012\left(\frac{\pi}{9}\right)\left[ \log|\sec x| \Big|_0^{\pi/9} + \frac{1}{2}\log|\sec 2x| \Big|_0^{\pi/9} + \frac{1}{3}\log|\sec 3x| \Big|_0^{\pi/9} + 0 \right]\] Now evaluate the difference at the limits: \[12 - 1012\left(\frac{\pi}{9}\right)\left[ (\log|\sec(\pi/9)| - \log|\sec0|) + \frac{1}{2}(\log|\sec(2\pi/9)| - \log|\sec0|) + \frac{1}{3}(\log|\sec(\pi/3)| - \log|\sec0|) \right]\] Evaluate the secant values and simplify the logarithms: \(12 - 1012\left(\frac{\pi}{9}\right)\left[ \log2 + \frac{1}{2}\log2 + \frac{1}{3}(\log2 + \log3 - \log1) \right]\) Combine all logarithm terms: \[12 - 1012\left(\frac{\pi}{9}\right)\left[\log2\left(1 + \frac{1}{2}\right) + \frac{1}{3}\log6\right]\] \[12 - 1128\log2 - 376\log3\] Now check the options: (a) \(\frac{1}{3}\log2 \approx 0.23\) (b) \(\log ^{3} \sqrt{4} \approx 0.602\) (c) \(3\log2 \approx 2.08\) (d) \(4\log\sqrt{3} \approx 2.191\) Our result, \(12 - 1128\log2 - 376\log3\), does not match any of the given options. Therefore, the problem might be ill-posed or we used an incorrect technique. Either way, the answer cannot be determined from the given information.

Key concepts

Integration Techniques

Integration is a powerful mathematical tool used for finding the antiderivative or area under a curve of a given function. In calculus, integration techniques consist of several methods used to simplify complex integrals and make them solvable. These techniques include:

• Substitution Method: This method is used when an integral can be transformed into a simpler one by using a substitution variable.


• Integration by Parts: This technique is based on the product rule for differentiation and is useful when the integrand is a product of two functions.


• Partial Fraction Decomposition: Often applied to rational functions, this technique breaks down the integrand into simpler fractions that can be integrated separately.


• Trigonometric Identities: Involves using identities to simplify the integration of trigonometric functions like sine, cosine, etc.

In the given exercise, the integration is broken down by separating terms, allowing the evaluation of each integral separately. This is done using the properties of integrals, where the integral of a sum is the sum of the integrals.

Definite Integrals

Definite integrals are used to calculate the total area under a curve, from one point to another on the x-axis. This area is not just a number, but the difference in the values of the antiderivative at these two points. A definite integral from point \(a\) to \(b\) can be expressed as:
\[\int_a^b f(x) \, dx = F(b) - F(a)\]where \(F(x)\) is the antiderivative of \(f(x)\). The notation \(F(b) - F(a)\) indicates calculating the upper and lower limits of the definite integral.
In this problem, we compute the definite integral of a complicated expression involving tangents over the interval \([0, \pi/9]\). Evaluating a definite integral accurately requires addressing each part separately, which is essential for handling integrals of complex trigonometric forms. By splitting into individual integrals, these can be solved individually and substituted back for a precise solution.

Trigonometric Functions

Understanding trigonometric functions is crucial in calculus, especially when integrating expressions that involve them. Common trigonometric functions include sine \(\sin(x)\), cosine \(\cos(x)\), and tangent \(\tan(x)\). Each function has specific properties and identities that are helpful for integration.
One important trigonometric identity used here is the tangent triple angle formula:
\[\tan 3x = 3\tan x - \tan^3 x\]This identity simplifies the product of tangent terms for easier integration. Additionally, expressions like \(\tan x \cdot \tan 2x \cdot \tan 3x\) require such identities for simplification before any integration attempt. Knowing how to manipulate these expressions effectively is vital for solving integrals involving complex trigonometric forms, as seen in this exercise.

Antiderivative Formulas

Antiderivative formulas, or indefinite integrals, are the reverse process of differentiation. They provide a function whose derivative equals the original function. Common antiderivative formulas include:

• \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\) for \(n eq -1\)
• \(\int \sin x \, dx = -\cos x + C\)
• \(\int \cos x \, dx = \sin x + C\)
• \(\int \tan x \, dx = \log|\sec x| + C\)

Applying these formulas correctly is essential for evaluating integrals, as seen in this exercise, where the integral of tangent functions was computed. Each formula helps to evaluate the integral over a specific interval, giving the result needed for definite integral computations. Understanding the direct link between derivatives and integrals through these formulas allows us to solve a broad range of calculus problems. These connections simplify complex integrations, making them approachable with systematic steps.