Evaluate sin integral (4x) cos(5x) dx

Calculus: Early Transcendentals
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Author:James Stewart
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Chapter1: Functions And Models
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Evaluate sin integral (4x) cos(5x) dx

The image contains a mathematical expression related to integral calculus involving trigonometric functions. Here is the transcription of the content as it would appear on an educational website:

---

### Example Problem: Integral of Trigonometric Functions

Solve the following integral:

\[ \int \sin(4x) \cos(5x) \, dx \]

#### Explanation:

To solve this integral, we can use trigonometric identities and integration techniques. In particular, we'll make use of the product-to-sum identities for sine and cosine functions.

One of the product-to-sum identities is:

\[ \sin(A) \cos(B) = \frac{1}{2} [ \sin(A + B) + \sin(A - B) ] \]

Applying this identity to \(\sin(4x) \cos(5x)\):

\[ \sin(4x) \cos(5x) = \frac{1}{2} [ \sin(4x + 5x) + \sin(4x - 5x) ] \]
\[ = \frac{1}{2} [ \sin(9x) + \sin(-x) ] \]

Since \(\sin(-x) = -\sin(x)\), the expression simplifies to:

\[ \sin(4x) \cos(5x) = \frac{1}{2} [ \sin(9x) - \sin(x) ] \]

Now, the integral becomes:

\[ \int \sin(4x) \cos(5x) \, dx = \int \frac{1}{2} [ \sin(9x) - \sin(x) ] \, dx \]
\[ = \frac{1}{2} \left[ \int \sin(9x) \, dx - \int \sin(x) \, dx \right] \]

The integrals can be evaluated using basic integration rules for sine functions:

\[ \int \sin(kx) \, dx = -\frac{1}{k} \cos(kx) + C \]

Thus:

\[ \int \sin(9x) \, dx = -\frac{1}{9} \cos(9x) \]
\[ \int \sin(x) \, dx = -\cos(x) \]

Substituting these results back into our expression:

\[ \frac{1}{2} \left[ -\
Transcribed Image Text:The image contains a mathematical expression related to integral calculus involving trigonometric functions. Here is the transcription of the content as it would appear on an educational website: --- ### Example Problem: Integral of Trigonometric Functions Solve the following integral: \[ \int \sin(4x) \cos(5x) \, dx \] #### Explanation: To solve this integral, we can use trigonometric identities and integration techniques. In particular, we'll make use of the product-to-sum identities for sine and cosine functions. One of the product-to-sum identities is: \[ \sin(A) \cos(B) = \frac{1}{2} [ \sin(A + B) + \sin(A - B) ] \] Applying this identity to \(\sin(4x) \cos(5x)\): \[ \sin(4x) \cos(5x) = \frac{1}{2} [ \sin(4x + 5x) + \sin(4x - 5x) ] \] \[ = \frac{1}{2} [ \sin(9x) + \sin(-x) ] \] Since \(\sin(-x) = -\sin(x)\), the expression simplifies to: \[ \sin(4x) \cos(5x) = \frac{1}{2} [ \sin(9x) - \sin(x) ] \] Now, the integral becomes: \[ \int \sin(4x) \cos(5x) \, dx = \int \frac{1}{2} [ \sin(9x) - \sin(x) ] \, dx \] \[ = \frac{1}{2} \left[ \int \sin(9x) \, dx - \int \sin(x) \, dx \right] \] The integrals can be evaluated using basic integration rules for sine functions: \[ \int \sin(kx) \, dx = -\frac{1}{k} \cos(kx) + C \] Thus: \[ \int \sin(9x) \, dx = -\frac{1}{9} \cos(9x) \] \[ \int \sin(x) \, dx = -\cos(x) \] Substituting these results back into our expression: \[ \frac{1}{2} \left[ -\
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