¡Claro! Vamos a resolver cada integral paso a paso: --- ### 1. \(\int \ln x^3\;dx\) **Paso 1:** Usa la propiedad de logaritmos: \(\ln(x^3) = 3\ln(x)\). \[ \int \ln x^3\; dx = \int 3\ln(x)\; dx = 3\int \ln(x)\; dx \] **Paso 2:** Usa integración por partes: Sea \(u = \ln(x)\), \(dv = dx\). Entonces, \(du = \frac{1}{x}dx\), \(v = x\). \[ \int \ln(x)\; dx = x\ln(x) - \int x \cdot \frac{1}{x} dx = x\ln(x) - \int 1 dx = x\ln(x) - x \] **Paso 3:** Multiplica por 3: \[ 3\int \ln(x)dx = 3[x\ln(x) - x] = 3x\ln(x) - 3x + C \] **Respuesta final:** \[ \boxed{3x\ln(x) - 3x + C} \] --- ### 2. \(\int (t^3 - 2t^2 + 4t - 3)e^{2t} dt\) Usa integración por partes repetidas, pero es más rápido usar la fórmula: \[ \int P(t) e^{at} dt = e^{at} Q(t) + C \] Donde \(Q(t)\) es un polinomio del mismo grado que \(P(t)\). **O bien**, usa la sustitución por partes repetidas (aquí el método tabular funciona muy bien): \[ \int P(t) e^{2t} dt = e^{2t} \left[ \frac{P(t)}{2} - \frac{P'(t)}{4} + \frac{P''(t)}{8} - \frac{P'''(t)}{16} \right] + C \] donde \(P(t) = t^3 - 2t^2 + 4t - 3\). Calculamos derivadas: - \(P'(t) = 3t^2 - 4t + 4\) - \(P''(t) = 6t - 4\) - \(P'''(t) = 6\) Entonces: \[ = e^{2t} \left( \frac{t^3 - 2t^2 + 4t - 3}{2} - \frac{3t^2 - 4t + 4}{4} + \frac{6t - 4}{8} - \frac{6}{16} \right) + C \] Simplifica: \[ = e^{2t} \left( \frac{t^3 - 2t^2 + 4t - 3}{2} - \frac{3t^2 - 4t + 4}{4} + \frac{3t - 2}{4} - \frac{3}{8} \right) + C \] \[ = e^{2t} \left( \frac{t^3}{2} - t^2 + 2t - \frac{3}{2} - \frac{3t^2}{4} + t - 1 + \frac{3t}{4} - \frac{1}{2} - \frac{3}{8} \right) + C \] Agrupa términos: \[ = e^{2t} \left( \frac{t^3}{2} - t^2 - \frac{3t^2}{4} + 2t + t + \frac{3t}{4} - \frac{3}{2} - 1 - \frac{1}{2} - \frac{3}{8} \right) + C \] \[ = e^{2t} \left( \frac{t^3}{2} - \frac{7t^2}{4} + \frac{15t}{4} - \frac{19}{8} \right) + C \] **Respuesta final:** \[ \boxed{e^{2t} \left( \frac{t^3}{2} - \frac{7t^2}{4} + \frac{15t}{4} - \frac{19}{8} \right) + C} \] --- ### 3. \(\int \sin^3x \cos^2x\; dx\) Vamos a usar sustitución: \[ \sin^3x \cos^2x = \sin x (1 - \cos^2 x)\cos^2 x = \sin x (\cos^2 x - \cos^4 x) \] Dejando \(u = \cos x\), \(du = -\sin x dx\): \[ \int \sin x (\cos^2 x - \cos^4 x) dx = -\int (u^2 - u^4) du = -\left( \frac{u^3}{3} - \frac{u^5}{5} \right) + C \] \[ = -\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C \] **Respuesta final:** \[ \boxed{-\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C} \] --- ### 4. \(\int \sec^3 \frac{x}{2} \tan^3 \frac{x}{2}\; dx\) Usa la relación \(\tan^2 u = \sec^2 u - 1\): \[ \tan^3 u = \tan u (\tan^2 u) = \tan u (\sec^2 u - 1) = \tan u \sec^2 u - \tan u \] Entonces, la integral es: \[ \int \sec^3 u \tan^3 u du = \int \sec^3 u (\tan u \sec^2 u - \tan u) du \] \[ = \int \sec^5 u \tan u du - \int \sec^3 u \tan u du \] Ahora, si \(u = \frac{x}{2}\), entonces \(du = \frac{1}{2}dx\), \(dx = 2 du\): Entonces la integral original se convierte en: \[ \int \sec^3 \frac{x}{2} \tan^3 \frac{x}{2} dx = 2 \int \sec^3 u \tan^3 u du \] Ahora, resolvemos cada parte: 1. \(\int \sec^5 u \tan u du\) - Usa sustitución: \(w = \sec u\), \(dw = \sec u \tan u du\). - Entonces, \(\tan u \sec^5 u du = \sec^4 u d(\sec u) = w^4 dw\). - \(\int w^4 dw = \frac{w^5}{5} = \frac{\sec^5 u}{5}\) 2. \(\int \sec^3 u \tan u du\) - Igual, \(w = \sec u\), \(dw = \sec u \tan u du\). - \(\int \sec^3 u \tan u du = \int \sec^2 u d(\sec u) = \frac{\sec^3 u}{3}\) Entonces, el resultado será: \[ 2 \left( \frac{\sec^5 u}{5} - \frac{\sec^3 u}{3} \right) + C \] \[ = \frac{2}{5} \sec^5 \frac{x}{2} - \frac{2}{3} \sec^3 \frac{x}{2} + C \] **Respuesta final:** \[ \boxed{\frac{2}{5} \sec^5 \frac{x}{2} - \frac{2}{3} \sec^3 \frac{x}{2} + C} \] --- ### 5. \(\int \sin 5x \cos 3x \; dx\) Usa la fórmula del producto a suma: \[ \sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)] \] Entonces, \[ \sin 5x \cos 3x = \frac{1}{2} [\sin(8x) + \sin(2x)] \] Integramos: \[ \int \sin 5x \cos 3x dx = \frac{1}{2} \int [\sin(8x) + \sin(2x)] dx \] \[ = \frac{1}{2} \left( -\frac{\cos 8x}{8} - \frac{\cos 2x}{2} \right) + C \] \[ = -\frac{1}{16} \cos 8x - \frac{1}{4} \cos 2x + C \] **Respuesta final:** \[ \boxed{-\frac{1}{16} \cos 8x - \frac{1}{4} \cos 2x + C} \] --- ¿Te gustaría ver la explicación detallada de alguna en específico?

Claro, aquí tienes las respuestas finales: 1. \( 3x \ln(x) - 3x + C \) 2. \( e^{2t} \left( \frac{t^3}{2} - \frac{7t^2}{4} + \frac{15t}{4} - \frac{19}{8} \right) + C \) 3. \( -\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C \) 4. \( \frac{2}{5} \sec^5 \frac{x}{2} - \frac{2}{3} \sec^3 \frac{x}{2} + C \) 5. \( -\frac{1}{16} \cos 8x - \frac{1}{4} \cos 2x + C \)

Certainly! Here's a detailed step-by-step solution for each integral, along with the final answers, all in English: --- ### 1. \(\int \ln x^3\, dx\) **Step 1:** Use the property of logarithms: \[ \ln x^3 = 3 \ln x \] So, the integral becomes: \[ \int 3 \ln x\, dx = 3 \int \ln x\, dx \] **Step 2:** Recall the standard integral: \[ \int \ln x\, dx \] Use integration by parts: - Let \(u = \ln x\) \(\Rightarrow du = \frac{1}{x} dx\) - Let \(dv = dx\) \(\Rightarrow v = x\) Applying integration by parts: \[ \int \ln x\, dx = x \ln x - \int x \cdot \frac{1}{x} dx = x \ln x - \int 1\, dx = x \ln x - x \] **Step 3:** Multiply by 3: \[ 3 \times (x \ln x - x) = 3x \ln x - 3x + C \] **Final answer:** \[ \boxed{3x \ln x - 3x + C} \] --- ### 2. \(\int (t^3 - 2t^2 + 4t - 3) e^{2t}\, dt\) This integral involves a polynomial times an exponential. To solve it efficiently, we'll use the method of repeated integration by parts or the tabular method for such products. **Step 1:** Let \[ P(t) = t^3 - 2t^2 + 4t - 3 \] The integral is: \[ \int P(t) e^{2t} dt \] **Step 2:** Use the tabular method. - Derive \(P(t)\) repeatedly: | Derivative Level | Polynomial | Result | |------------------|--------------|---------------------------| | 0 (original) | \(t^3 - 2t^2 + 4t - 3\) | \(P(t)\) | | 1 | \(3t^2 - 4t + 4\) | \(P'(t)\) | | 2 | \(6t - 4\) | \(P''(t)\) | | 3 | 6 | \(P'''(t)\) | - Integrate \(e^{2t}\) repeatedly: \[ \int e^{2t}\, dt = \frac{1}{2} e^{2t} \] - The tabular method gives: \[ \int P(t) e^{2t}\, dt = e^{2t} \times \left( \frac{P(t)}{2} - \frac{P'(t)}{4} + \frac{P''(t)}{8} - \frac{P'''(t)}{16} \right) + C \] **Step 3:** Plug in the derivatives: \[ = e^{2t} \left[ \frac{t^3 - 2t^2 + 4t - 3}{2} - \frac{3t^2 - 4t + 4}{4} + \frac{6t - 4}{8} - \frac{6}{16} \right] + C \] **Step 4:** Simplify each term: \[ = e^{2t} \left( \frac{t^3 - 2t^2 + 4t - 3}{2} - \frac{3t^2 - 4t + 4}{4} + \frac{6t - 4}{8} - \frac{6}{16} \right) \] Express all with denominator 8: \[ = e^{2t} \left( \frac{4(t^3 - 2t^2 + 4t - 3)}{8} - \frac{2(3t^2 - 4t + 4)}{8} + \frac{6t - 4}{8} - \frac{6}{16} \right) \] Note that \(\frac{6}{16} = \frac{3}{8}\). Now, write all over denominator 8: \[ = e^{2t} \left( \frac{4t^3 - 8t^2 + 16t - 12 - 6t^2 + 8t - 8 + 6t - 4 - 3}{8} \right) \] Combine like terms numerator: - \(t^3\): \(4t^3\) - \(t^2\): \(-8t^2 - 6t^2 = -14t^2\) - \(t\): \(16t + 8t + 6t = 30t\) - Constants: \(-12 - 8 - 4 - 3 = -27\) **Final simplified form:** \[ e^{2t} \times \frac{4t^3 - 14 t^2 + 30 t - 27}{8} + C \] **Final answer:** \[ \boxed{e^{2t} \left( \frac{4t^3 - 14 t^2 + 30 t - 27}{8} \right) + C} \] --- ### 3. \(\int \sin^3 x \cos^2 x\, dx\) **Step 1:** Express \(\sin^3 x\): \[ \sin^3 x = \sin x \cdot \sin^2 x \] Recall: \[ \sin^2 x = 1 - \cos^2 x \] Thus: \[ \sin^3 x = \sin x (1 - \cos^2 x) \] The integral becomes: \[ \int \sin x (1 - \cos^2 x) \cos^2 x\, dx \] But to simplify, it's easier to substitute: \[ u = \cos x \quad \Rightarrow du = -\sin x dx \] Rearranged: \[ \sin x dx = - du \] Express \(\sin^3 x \cos^2 x\, dx\): \[ = \sin x \cdot \sin^2 x \cdot \cos^2 x\, dx \] Replace \(\sin^2 x = 1 - u^2\): \[ = \sin x (1 - u^2) u^2 dx \] But since \(\sin x dx = - du\), and \(\sin x dx\) appears once: \[ \Rightarrow \sin x dx = - du \] The integral becomes: \[ \int (1 - u^2) u^2 \times \sin x dx = - \int (1 - u^2) u^2 du \] Expand numerator: \[ - \int (u^2 - u^4) du \] Integrate: \[ - \left( \frac{u^3}{3} - \frac{u^5}{5} \right) + C \] Replace back \(u = \cos x\): \[ - \left( \frac{\cos^3 x}{3} - \frac{\cos^5 x}{5} \right) + C \] **Final answer:** \[ \boxed{ - \frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C} \] --- ### 4. \(\int \sec^3 \frac{x}{2} \tan^3 \frac{x}{2}\, dx\) **Step 1:** Rewrite the integral: \[ \int \sec^3 u \tan^3 u\, dx \] where \(u = \frac{x}{2}\), so: \[ du = \frac{1}{2} dx \Rightarrow dx = 2 du \] Hence, the integral becomes: \[ 2 \int \sec^3 u \tan^3 u\, du \] **Step 2:** Express \(\tan^3 u\): \[ \tan^3 u = \tan u (\tan^2 u) = \tan u (\sec^2 u - 1) \] So: \[ 2 \int \sec^3 u \tan u (\sec^2 u - 1)\, du \] Distribute: \[ 2 \int \sec^3 u \tan u \sec^2 u\, du - 2 \int \sec^3 u \tan u\, du \] Notice: \[ \sec^3 u \tan u \sec^2 u = \sec^5 u \tan u \] and \[ \sec^3 u \tan u \] So, the integral is: \[ 2 \left( \int \sec^5 u \tan u\, du - \int \sec^3 u \tan u\, du \right) \] **Step 3:** Use substitution: - For \(\int \sec^n u \tan u\, du\), let \(w = \sec u\), \(dw = \sec u \tan u\, du\). Then: \[ \sec u \tan u\, du = dw \Rightarrow \tan u\, du = \frac{dw}{w} \] But in our integrals, the integrand is: \[ \sec^n u \tan u\, du = w^n \tan u\, du \] Substituting: \[ w^n \times \frac{dw}{w} = w^{n-1} dw \] Therefore: \[ \int \sec^n u \tan u\, du = \int w^{n-1} dw = \frac{w^{n}}{n} \] **Step 4:** Apply to both integrals: - For \(n=5\): \[ \int \sec^5 u \tan u\, du = \frac{\sec^5 u}{5} \] - For \(n=3\): \[ \int \sec^3 u \tan u\, du = \frac{\sec^3 u}{3} \] Putting it all together: \[ 2 \left( \frac{\sec^5 u}{5} - \frac{\sec^3 u}{3} \right) + C \] Recall \(u = \frac{x}{2}\): \[ \boxed{\frac{2}{5} \sec^5 \frac{x}{2} - \frac{2}{3} \sec^3 \frac{x}{2} + C} \] --- ### 5. \(\int \sin 5x \cos 3x\, dx\) **Step 1:** Use product-to-sum identities: \[ \sin A \cos B = \frac{1}{2} [\sin(A+B) + \sin(A-B)] \] So: \[ \sin 5x \cos 3x = \frac{1}{2} [\sin(8x) + \sin(2x)] \] **Step 2:** Integrate term-by-term: \[ \int \sin 5x \cos 3x\, dx = \frac{1}{2} \left( \int \sin 8x\, dx + \int \sin 2x\, dx \right) \] The integrals: \[ \int \sin kx\, dx = - \frac{\cos kx}{k} \] Thus: \[ = \frac{1}{2} \left( - \frac{\cos 8x}{8} - \frac{\cos 2x}{2} \right) + C \] Simplify: \[ = - \frac{\cos 8x}{16} - \frac{\cos 2x}{4} + C \] **Final answer:** \[ \boxed{ - \frac{\cos 8x}{16} - \frac{\cos 2x}{4} + C} \] --- **Summary of Final Answers:** 1. \(\boxed{3x \ln x - 3x + C}\) 2. \(\boxed{e^{2t} \left( \frac{t^3}{2} - \frac{7 t^2}{4} + \frac{15 t}{4} - \frac{19}{8} \right) + C}\) 3. \(\boxed{- \frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C}\) 4. \(\boxed{\frac{2}{5} \sec^5 \frac{x}{2} - \frac{2}{3} \sec^3 \frac{x}{2} + C}\) 5. \(\boxed{- \frac{\cos 8x}{16} - \frac{\cos 2x}{4} + C}\) Let me know if you'd like further clarifications!