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Evaluate the integral. $$\int e^{3 x} \cos 2 x d x$$

Short Answer

Expert verified
The integral is \( \frac{3}{13} e^{3x} \cos 2x + \frac{2}{13} e^{3x} \sin 2x + C \).

Step by step solution

01

Identify the Integral Type

The given integral \( \int e^{3x} \cos 2x \, dx \) involves both an exponential function and a trigonometric function. This type of integral is typically solved using the method of integration by parts, applied twice.
02

Apply Integration by Parts

Use the integration by parts formula \( \int u \, dv = uv - \int v \, du \). For the first application, let \( u = \cos 2x \) and \( dv = e^{3x} \, dx \). Then \( du = -2 \sin 2x \, dx \) and \( v = \frac{1}{3} e^{3x} \). Apply the formula:\[ \int e^{3x} \cos 2x \, dx = \frac{1}{3} e^{3x} \cos 2x - \int \frac{1}{3} e^{3x} (-2 \sin 2x) \, dx \]
03

Simplify the First Result

Simplify the result from Step 2:\[ \frac{1}{3} e^{3x} \cos 2x + \frac{2}{3} \int e^{3x} \sin 2x \, dx \]The integral remaining is \( \int e^{3x} \sin 2x \, dx \), which requires applying integration by parts again.
04

Apply Integration by Parts Again

For \( \int e^{3x} \sin 2x \, dx \), let \( u = \sin 2x \) and \( dv = e^{3x} \, dx \). Then \( du = 2 \cos 2x \, dx \) and \( v = \frac{1}{3} e^{3x} \). Apply integration by parts:\[ \int e^{3x} \sin 2x \, dx = \frac{1}{3} e^{3x} \sin 2x - \int \frac{1}{3} e^{3x} \cdot 2 \cos 2x \, dx \]
05

Substitute and Solve for Original Integral

Substitute the result back into the original expression:\[ \int e^{3x} \sin 2x \, dx = \frac{1}{3} e^{3x} \sin 2x - \frac{2}{3} \int e^{3x} \cos 2x \, dx \]Substitute this back:\[ \frac{1}{3} e^{3x} \cos 2x + \frac{2}{3} \left( \frac{1}{3} e^{3x} \sin 2x - \frac{2}{3} \int e^{3x} \cos 2x \, dx \right) = \int e^{3x} \cos 2x \, dx \]
06

Simplify and Solve for the Integral

Simplify the equation:\[ \frac{1}{3} e^{3x} \cos 2x + \frac{2}{9} e^{3x} \sin 2x - \frac{4}{9} \int e^{3x} \cos 2x \, dx = \int e^{3x} \cos 2x \, dx \]Rearrange to get:\[ \int e^{3x} \cos 2x \, dx + \frac{4}{9} \int e^{3x} \cos 2x \, dx = \frac{1}{3} e^{3x} \cos 2x + \frac{2}{9} e^{3x} \sin 2x \]
07

Solve for the Integral

Combine terms of the integral:\[ \frac{13}{9} \int e^{3x} \cos 2x \, dx = \frac{1}{3} e^{3x} \cos 2x + \frac{2}{9} e^{3x} \sin 2x \]Divide both sides by \( \frac{13}{9} \) to solve for \( \int e^{3x} \cos 2x \, dx \):\[ \int e^{3x} \cos 2x \, dx = \frac{9}{13} \left( \frac{1}{3} e^{3x} \cos 2x + \frac{2}{9} e^{3x} \sin 2x \right) \]
08

Simplify the Final Answer

Distribute \( \frac{9}{13} \):\[ \int e^{3x} \cos 2x \, dx = \frac{3}{13} e^{3x} \cos 2x + \frac{2}{13} e^{3x} \sin 2x + C \] where \( C \) is the integration constant.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Functions
Trigonometric functions are fundamental in calculus and mathematics, generally related to angles and their side ratios in a right triangle. The primary trigonometric functions include sine (\( \sin \theta \)), cosine (\( \cos \theta \)), and tangent (\( \tan \theta \)). These functions are periodic, with sine and cosine having a period of \( 2\pi \), and tangent having a period of \( \pi \).

In the given integral \( \int e^{3x} \cos 2x \, dx \), the function involved is the cosine function. Cosine is known for its wave-like nature and is often compared and combined with other functions, like exponential ones, especially in calculus.

Key points to remember about trigonometric functions include:
  • Sine and cosine are used to describe oscillatory motion, like waves.
  • These functions are differentiable, with derivatives and integrals frequently used in solving calculus problems.
  • Cosine's derivative is negative sine, while sine's derivative is cosine. Recognizing these derivatives helps in solving differential integrals involving trigonometric functions.
Exponential Functions
Exponential functions are another crucial component of calculus, characterized by the form \( f(x) = a^{x} \), where the base \( a \) is a constant and often taken as \( e \), the mathematical constant approximately equal to 2.71828. Exponential functions grow rapidly, making them ideal for modeling processes like population growth or radioactive decay.

In the integral \( \int e^{3x} \cos 2x \, dx \), \( e^{3x} \) is the exponential function. Understanding the properties of exponential functions is key to solving integrals by implementing rule-based strategies in calculus like integration by parts.

Notable features of exponential functions include:
  • They are continuously increasing or decreasing, depending on the exponent's sign.
  • The derivative and integral of \( e^{x} \) are unique in that they both remain \( e^{x} \), but when coefficients are involved, like \( 3x \), these influence scaling factors.
  • When combined with trigonometric functions, exponential functions often require advanced integration techniques, such as integration by parts, to simplify and solve.
Definite and Indefinite Integrals
Integrals are one of the central concepts in calculus, representing the signed area under a curve. There are two main types: definite integrals and indefinite integrals. Each serves different purposes and is used in distinct scenarios.

An indefinite integral, like \( \int e^{3x} \cos 2x \, dx \), is a function representation, often including an integration constant \( C \), which accounts for any unknown vertical shifts. Indefinite integrals are used for finding antiderivatives, revealing all possible functions that could result from differentiation.

Definite integrals, on the other hand, compute the actual area under the curve between two endpoints, resulting in a numerical value. They are less concerned with general antiderivatives and more with quantifying accumulated quantities.

Hints for approaching integrals:
  • For indefinite integrals, always include the integration constant \( C \) unless boundary conditions are specified.
  • Understanding the function type within an integral, whether exponential, trigonometric, or polynomial, helps in choosing the right integration technique, such as substitution or integration by parts.
  • The method of integration by parts is particularly handy when dealing with products of functions, such as the combination of \( e^{3x} \) and \( \cos 2x \) in this case.

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Most popular questions from this chapter

(a) In the integral \(\int x \cos x \, d x,\) let $$ \begin{array}{l} u=x, \quad d v=\cos x d x \\ d u=d x, \quad v=\sin x+C_{1}\end{array}$$ Show that the constant \(C_{1}\) cancels out, thus giving the same solution obtained by omitting \(C_{1}\) (b) Show that in general \(u v-\int v \, d u=u\left(v+C_{1}\right)-\int\left(v+C_{1}\right) d u$$ thereby justifying the omission of the constant of integration when calculating \)v$ in integration by parts.

Evaluate the integral. $$\int_{0}^{1 / 4} \sec \pi x \tan \pi x d x$$

Express the improper integral as a limit, and then evaluate that limit with a CAS. Confirm the answer by evaluating the integral directly with the CAS. $$\int_{0}^{+\infty} e^{-x} \cos x d x$$

Use any method to find the area of the region enclosed by the curves. $$y=\frac{1}{25-16 x^{2}}, y=0, x=0, x=1$$

A transform is a formula that converts or "transforms" one function into another. Transforms are used in applications to convert a difficult problem into an easier problem whose solution can then be used to solve the original difficult problem. The Laplace transform of a function \(f(t),\) which plays an important role in the study of differential equations, is denoted by \(\mathscr{L}\\{f(t)\\}\) and is defined by $$\mathscr{L}\\{f(t)\\}=\int_{0}^{+\infty} e^{-s t} f(t) d t$$ In this formula \(s\) is treated as a constant in the integration process; thus, the Laplace transform has the effect of transforming \(f(t)\) into a function of \(s .\) Use this formula in these exercises. In each part, find the Laplace transform. (a) \(f(t)=t, s>0\) (b) \(f(t)=t^{2}, s>0\) (c) \(f(t)=\left\\{\begin{array}{ll}0, & t<3 \\ 1, & t \geq 3\end{array}, s>0\right.\)

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