Chapter 16: Chapter 16

Problem 10

Evaluate the integrals. \(\int x \cos \left(-3 x^{2}+4\right) d x\)

Short Answer

Expert verified

The solution to the integral \(\int x \cos \left(-3 x^{2}+4\right) d x\) is \(-\dfrac{1}{6} \sin(-3x^2+4)+C\), where \(C\) is the constant of integration.

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Step by step solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Identify the substitution

We will substitute \(u = -3x^2 + 4\). Observe that \(du = -6x dx\), which can be easily found by taking the derivative of \(u\) with respect to \(x\).

Step 2: Rewrite the integral in terms of the new variable

Now that we have our substitution, let's rewrite the integral in terms of \(u\): To isolate \(dx\), we will divide both sides by \(-6x\): \(dx = \frac{du}{-6x}\) Now, let's replace the original expression with the one in terms of \(u\): \(\int x \cos(-3x^2+4) dx = \int x \cos(u) \left(\frac{du}{-6x}\right)\) Notice that the \(x\) terms cancel out: \(\int \cos(u) \left(\frac{du}{-6}\right)\)

Step 3: Integrate with respect to the new variable

Now that we have an expression that is simpler to integrate, let's perform the integration: \(-\dfrac{1}{6}\int \cos(u)du\) The integral of \(\cos(u)\) is \(\sin(u)\), so: \(-\dfrac{1}{6} \sin(u) + C_1\) (where \(C_1\) is the constant of integration)

Step 4: Substitute the original variable back to get the final result

Now, let's substitute the original variable back by replacing \(u\) with \(-3x^2 + 4\). So, the final expression becomes: \(-\dfrac{1}{6} \sin(-3x^2+4)+C\) Where \(C\) is the constant of integration. Therefore, the solution to the integral is: \(\int x \cos \left(-3 x^{2}+4\right) d x = -\dfrac{1}{6} \sin(-3x^2+4)+C\)

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Chapter 16: Chapter 16

Evaluate the integrals. \(\int x \cos \left(-3 x^{2}+4\right) d x\)

Short Answer

Step by step solution

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Step 1: Identify the substitution

Step 2: Rewrite the integral in terms of the new variable

Step 3: Integrate with respect to the new variable

Step 4: Substitute the original variable back to get the final result

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More chapters from the book ‘Finite Mathematics and Applied Calculus’

Chapter 0

Chapter 1

Chapter 2

Chapter 3

Chapter 4

Chapter 5

Chapter 6

Chapter 7

Chapter 8

Chapter 9

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 14

Chapter 15

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