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\( \int \tan (5 x) \sec ^{2}(5 x) d x \) Seleccione una: a. \( \frac{1}{10} \tan ^{2}(5 x)+C \) b. \( -\frac{1}{10} \sec ^{2}(5 x)+C \) c. \( 10 \sec ^{3}(5 x)+C \) d. \( \frac{1}{10} \tan ^{2}(5 x)+\frac{1}{10} \sec (5 x)+C \)

Ask by Rodriguez Schultz. in Mexico
Mar 13,2025

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Answer

La integral \( \int \tan (5 x) \sec ^{2}(5 x) d x \) se resuelve al usar la sustitución \( u = \tan(5x) \), lo que da como resultado \( \frac{1}{10} \tan ^{2}(5 x)+C \).

Solution

Para resolver la integral \( \int \tan (5 x) \sec ^{2}(5 x) d x \), podemos utilizar una sustitución. 1. **Identificamos la función a derivar**: Observamos que la derivada de \( \tan(5x) \) es \( 5 \sec^2(5x) \). Esto sugiere que podemos usar \( u = \tan(5x) \) como nuestra sustitución. 2. **Calculamos \( du \)**: \[ du = 5 \sec^2(5x) \, dx \implies dx = \frac{du}{5 \sec^2(5x)} \] 3. **Sustituimos en la integral**: \[ \int \tan(5x) \sec^2(5x) \, dx = \int u \cdot \sec^2(5x) \cdot \frac{du}{5 \sec^2(5x)} = \frac{1}{5} \int u \, du \] 4. **Resolvemos la integral**: \[ \frac{1}{5} \int u \, du = \frac{1}{5} \cdot \frac{u^2}{2} = \frac{u^2}{10} + C \] 5. **Volvemos a sustituir \( u \)**: \[ \frac{u^2}{10} + C = \frac{\tan^2(5x)}{10} + C \] Por lo tanto, la solución a la integral es: \[ \frac{1}{10} \tan^2(5x) + C \] La respuesta correcta es: **a. \( \frac{1}{10} \tan ^{2}(5 x)+C \)**.

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The Deep Dive

To solve the integral \( \int \tan(5x) \sec^2(5x) \, dx \), we can use a substitution. Let \( u = \tan(5x) \), then \( du = 5 \sec^2(5x) \, dx \) or \( dx = \frac{du}{5 \sec^2(5x)} \). This simplifies the integral to: \[ \int \tan(5x) \sec^2(5x) \, dx = \frac{1}{5} \int u \, du = \frac{1}{5} \cdot \frac{u^2}{2} = \frac{1}{10} u^2 + C = \frac{1}{10} \tan^2(5x) + C \] Based on this evaluation, the correct answer is: **a. \( \frac{1}{10} \tan^{2}(5x) + C \)** --- The derivative of \( \tan(x) \) is \( \sec^2(x) \), which is key here! This tie-in allows you to spot integrals that incorporate both functions easily. If you recognize these relationships, integrating trigonometric functions becomes second nature! Another interesting point: substitutions work best when you can quickly derive the substitute back to the original variables. Remember to differentiate your choices well, as it can lead to recognizing patterns more rapidly in complex expressions!

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