Math Problem Statement

Compute the integral of \( \sin^3(x) e^{\cos(x)} \cos(x) \, dx \) using substitution and integration by parts.

Solution

The problem focuses on solving the integral of sin3(x)ecos(x)cos(x)dx\sin^3(x)e^{\cos(x)} \cos(x) \, dx using substitution and integration by parts.

Key Steps:

  1. Substitution: Let u=cos(x)u = \cos(x), so du=sin(x)dxdu = -\sin(x) \, dx. This transforms the integral into terms of uu.

  2. Rewriting the Integral: The original integral becomes: sin3(x)ecos(x)cos(x)dx=(1u2)eudu\int \sin^3(x) e^{\cos(x)} \cos(x) \, dx = -\int (1 - u^2) e^u \, du

  3. Integration by Parts (First Time): Decompose the integral (1u2)eudu\int (1 - u^2)e^u \, du using: f(u)=(1u2),g(u)=eu,g(u)=eu.f(u) = (1 - u^2), \quad g'(u) = e^u, \quad g(u) = e^u.

    Apply the integration by parts formula: f(u)g(u)du=f(u)g(u)f(u)g(u)du\int f(u)g'(u) \, du = f(u)g(u) - \int f'(u)g(u) \, du Compute f(u)f'(u), then solve the resulting integrals step-by-step.

  4. Second Integration by Parts: For the resulting integral, use f(u)=2uf(u) = 2u and g(u)=eug'(u) = e^u, and compute similarly.

  5. Final Expression: The solution simplifies to: sin3(x)ecos(x)cos(x)dx=(cos(x)22cos(x)+1)ecos(x)+C\int \sin^3(x)e^{\cos(x)}\cos(x) \, dx = \left( \cos(x)^2 - 2\cos(x) + 1 \right)e^{\cos(x)} + C

Would you like me to delve deeper into any step, or help with verification? Let me know!


Related Questions:

  1. What is the significance of substitution in integral problems?
  2. How does integration by parts work conceptually?
  3. Why is sin(x)dx-\sin(x)dx critical when substituting u=cos(x)u = \cos(x)?
  4. Can we directly integrate sin3(x)ecos(x)cos(x)\sin^3(x)e^{\cos(x)} \cos(x)? Why or why not?
  5. How does the exponential function eue^u simplify integration by parts?

Tip: Always double-check your substitutions when solving complex integrals to avoid errors!

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Math Problem Analysis

Mathematical Concepts

Integration by substitution
Integration by parts
Exponential functions
Trigonometric identities

Formulas

Substitution: \( u = \cos(x), \; du = -\sin(x) \, dx \)
Integration by parts: \( \int f(u)g'(u) \, du = f(u)g(u) - \int f'(u)g(u) \, du \)

Theorems

Integration by substitution theorem
Integration by parts theorem

Suitable Grade Level

Undergraduate Calculus