Use integration by parts to evaluate the integral: [tex]$\int \cos (\ln (5 x)) d x$[/tex]
Answer (2)
The integral ∫ cos ( ln ( 5 x )) d x can be evaluated using integration by parts twice. The final result is 2 x ( cos ( ln ( 5 x )) + sin ( ln ( 5 x )) ) + C , where C is the constant of integration. This method provides insight into the integration process involving trigonometric and logarithmic functions.
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Apply integration by parts with u = cos ( ln ( 5 x )) and d v = d x .
Find d u = − x s i n ( l n ( 5 x )) d x and v = x .
Apply integration by parts again to ∫ sin ( ln ( 5 x )) d x , with u = sin ( ln ( 5 x )) and d v = d x .
Solve for the original integral to get 2 x ( cos ( ln ( 5 x )) + sin ( ln ( 5 x ))) + C .
Explanation
Problem Setup We are asked to evaluate the integral ∫ cos ( ln ( 5 x )) d x using integration by parts.
Choosing u and dv Let's use integration by parts. We'll set u = cos ( ln ( 5 x )) and d v = d x . Then we need to find d u and v .
Finding du First, let's find d u . The derivative of u = cos ( ln ( 5 x )) is d u = − sin ( ln ( 5 x )) ⋅ 5 x 1 ⋅ 5 d x = − x s i n ( l n ( 5 x )) d x .
Finding v Next, let's find v . Since d v = d x , integrating gives us v = x .
Applying Integration by Parts Now we apply the integration by parts formula: ∫ u d v = uv − ∫ v d u . So, ∫ cos ( ln ( 5 x )) d x = x cos ( ln ( 5 x )) − ∫ x ⋅ ( − x s i n ( l n ( 5 x )) ) d x = x cos ( ln ( 5 x )) + ∫ sin ( ln ( 5 x )) d x .
Integrating the Second Term Now we need to evaluate ∫ sin ( ln ( 5 x )) d x . We'll use integration by parts again. Let u = sin ( ln ( 5 x )) and d v = d x . Then d u = x c o s ( l n ( 5 x )) d x and v = x .
Applying Integration by Parts Again Applying integration by parts again: ∫ sin ( ln ( 5 x )) d x = x sin ( ln ( 5 x )) − ∫ x ⋅ x c o s ( l n ( 5 x )) d x = x sin ( ln ( 5 x )) − ∫ cos ( ln ( 5 x )) d x .
Substituting Back Substitute this result back into the original equation: ∫ cos ( ln ( 5 x )) d x = x cos ( ln ( 5 x )) + x sin ( ln ( 5 x )) − ∫ cos ( ln ( 5 x )) d x .
Rearranging the Equation Now, we rearrange the equation to solve for ∫ cos ( ln ( 5 x )) d x : 2 ∫ cos ( ln ( 5 x )) d x = x cos ( ln ( 5 x )) + x sin ( ln ( 5 x )) .
Final Result Finally, divide by 2: ∫ cos ( ln ( 5 x )) d x = 2 x ( cos ( ln ( 5 x )) + sin ( ln ( 5 x )) ) + C , where C is the constant of integration.
Examples
Imagine you're designing an audio filter where the frequency response is modeled by a cosine function of the logarithm of the frequency. Evaluating integrals like this helps you understand the filter's overall behavior and how it affects different frequency components in an audio signal. This is crucial in audio engineering for creating effects, noise reduction, and signal enhancement.
