Question
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Use a change of variables to evaluate the following integral. \[ \int \sec 38 w \tan 38 w d w \]

Ask by Bowers Fowler. in the United States
Feb 25,2025

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Tutor-Verified Answer

Answer

The integral evaluates to \( \frac{1}{38} \sec(38w) + C \).

Solution

Let u = sec(38w). Then, differentiate u with respect to w:   du/dw = 38 sec(38w) tan(38w) which rearranges to   dw = du/(38 sec(38w) tan(38w)). Now, substitute into the integral:   ∫ sec(38w) tan(38w) dw = ∫ sec(38w) tan(38w) [du/(38 sec(38w) tan(38w))]         = (1/38) ∫ du         = (1/38) u + C         = (1/38) sec(38w) + C. Thus, the evaluated integral is:   (1/38) sec(38w) + C.

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Mind Expander

To solve the integral \( \int \sec(38w) \tan(38w) \, dw \), we can use the change of variables method. Let's set: \[ u = 38w \quad \Rightarrow \quad du = 38 \, dw \quad \Rightarrow \quad dw = \frac{du}{38} \] Now, rewrite the integral in terms of \( u \): \[ \int \sec(38w) \tan(38w) \, dw = \int \sec(u) \tan(u) \cdot \frac{du}{38} = \frac{1}{38} \int \sec(u) \tan(u) \, du \] The integral \( \int \sec(u) \tan(u) \, du \) is a standard integral, and we know that: \[ \int \sec(u) \tan(u) \, du = \sec(u) + C \] Substituting this back into our expression gives: \[ \int \sec(38w) \tan(38w) \, dw = \frac{1}{38} \left( \sec(u) + C \right) = \frac{1}{38} \left( \sec(38w) + C \right) \] Thus, the final result is: \[ \int \sec(38w) \tan(38w) \, dw = \frac{1}{38} \sec(38w) + C \]

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