Problem 69
Question
Use the reduction formulas in a table of integrals to evaluate the following integrals. $$\int \tan ^{4} 3 y d y$$
Step-by-Step Solution
Verified Answer
Question: Determine the integral of the function \(f(y) = \tan^4 3y\).
Answer: The integral of the function \(f(y) = \tan^4 3y\) is given by \(\int \tan^4 3y dy = \frac{1}{3}\tan^2 3y \sec^2 3y - \frac{2}{9}\tan 3y + D\), where D is the constant of integration.
1Step 1: Identify the power and use the reduction formula for the first time
We have an integral of the form \(\int \tan^4 3y dy\), where n = 4. Applying the reduction formula, we get:
$$\int \tan^4 3y dy = \frac{1}{4 - 1}\tan^{4 - 2} 3y \sec^2 3y - \frac{4 - 2}{4 - 1} \int \tan^{4 - 2} 3y dy$$
$$\int \tan^4 3y dy = \frac{1}{3}\tan^2 3y \sec^2 3y - \frac{2}{3} \int \tan^2 3y dy$$
2Step 2: Reduce the power further using the reduction formula and simplify
We still have an integral involving a power of the tangent function, so we'll apply the reduction formula again to the remaining integral:
$$\int \tan^2 3y dy = \frac{1}{2 - 1}\tan^{2 - 2} 3y \sec^2 3y - \frac{2 - 2}{2 - 1} \int \tan^{2 - 2} 3y dy$$
$$\int \tan^2 3y dy = \tan^0 3y \sec^2 3y = \sec^2 3y$$
Now substitute this result back into our previous expression:
$$\int \tan^4 3y dy = \frac{1}{3}\tan^2 3y \sec^2 3y - \frac{2}{3} \int \sec^2 3y dy$$
3Step 3: Integrate the remaining term
We have the integral \(\int \sec^2 3y dy\) remaining, which can be directly integrated:
$$\int \sec^2 3y dy = \frac{1}{3} \tan 3y + C$$
Substitute this result back into our expression:
$$\int \tan^4 3y dy = \frac{1}{3}\tan^2 3y \sec^2 3y - \frac{2}{3}\left(\frac{1}{3}\tan 3y + C\right)$$
4Step 4: Simplify and write the final answer
Now we'll simplify the expression:
$$\int \tan^4 3y dy = \frac{1}{3}\tan^2 3y \sec^2 3y - \frac{2}{9}\tan 3y - \frac{2}{3}C$$
Let's denote the constant with a different letter, say \(D = \frac{2}{3}C\). So the final answer will be:
$$\int \tan^4 3y dy = \frac{1}{3}\tan^2 3y \sec^2 3y - \frac{2}{9}\tan 3y + D$$
Key Concepts
Integral CalculusPower of Tangent FunctionIntegration Techniques
Integral Calculus
Integral calculus is a major part of calculus concerned with finding the integral of a function. It helps in understanding quantities like area, volume, and accumulation.
There are two main types of integrals:
When solving integrals involving trigonometric functions, such as \(\int \tan^4 3y dy\), we often use techniques like substitution, integration by parts, or special formulas such as reduction formulas. These techniques transform complicated integrals into simpler forms that are easier to evaluate.
There are two main types of integrals:
- Definite Integrals: Provide a number representing the signed area under a curve and between specified limits.
- Indefinite Integrals: Represent a family of functions and include a constant of integration, often noted as +C.
When solving integrals involving trigonometric functions, such as \(\int \tan^4 3y dy\), we often use techniques like substitution, integration by parts, or special formulas such as reduction formulas. These techniques transform complicated integrals into simpler forms that are easier to evaluate.
Power of Tangent Function
The tangent function, denoted as \( \tan(x) \), is periodic and related to the sine and cosine functions. It is used frequently in trigonometry and calculus.
In integrals like \( \int \tan^4 3y \, dy \), we deal with high powers of the tangent function.
To work with the tangent function in calculus:
In integrals like \( \int \tan^4 3y \, dy \), we deal with high powers of the tangent function.
To work with the tangent function in calculus:
- Reduction Formula: For an integral of the form \( \int \tan^n(x)dx \), reduction formulas help break down higher powers into more manageable terms.
- Power Reduction Identities: Sometimes, trigonometric identities are used to reduce the power of functions, for example, expressing tangent in terms of sine and cosine.
Integration Techniques
Solving complex integrals often requires sophisticated techniques to simplify calculations. These techniques aim to transform a challenging integral into one that is straightforward to solve.
Here are some common techniques:
Here are some common techniques:
- Reduction Formulas: These are particularly useful for functions raised to a power, like \( \tan^n(x) \). They help diminish the power progressively.
- Substitution: This involves changing variables to simplify the integration. For instance, substituting \( u = 3y \) can sometimes help manage multipliers and exponents.
- Direct Integration: Often pursued after reducing the more complex parts to simpler integrals, as in the case of \( \int \sec^2(x)dx = \tan(x) + C \).
Other exercises in this chapter
Problem 69
Consider the general first-order initial value problem \(y^{\prime}(t)=a y+b, y(0)=y_{0},\) for \(t \geq 0,\) where \(a, b,\) and \(y_{0}\) are real numbers. a.
View solution Problem 69
Graph the integrands and then evaluate and compare the values of \(\int_{0}^{\infty} x e^{-x^{2}} d x\) and \(\int_{0}^{\infty} x^{2} e^{-x^{2}} d x\).
View solution Problem 69
Integrals of the form \(\int \sin m x \cos n x d x\) Use the following three identities to evaluate the given integrals. $$\begin{aligned}&\sin m x \sin n x=\fr
View solution Problem 69
Suppose a mass on a spring that is slowed by friction has the position function \(s(t)=e^{-t} \sin t\) a. Graph the position function. At what times does the os
View solution