Math Problem Statement

∫ π 0 esin(t) cos(t) sin(t)dt

Solution

We are asked to evaluate the definite integral:

I=0πesin(t)cos(t)sin(t)dtI = \int_0^\pi e^{\sin(t)} \cos(t) \sin(t) \, dt

To solve this, notice that we can rewrite the product of cos(t)sin(t)\cos(t) \sin(t) as:

cos(t)sin(t)=12sin(2t)\cos(t) \sin(t) = \frac{1}{2} \sin(2t)

Thus, the integral becomes:

I=0πesin(t)12sin(2t)dtI = \int_0^\pi e^{\sin(t)} \cdot \frac{1}{2} \sin(2t) \, dt

Now let's simplify this integral and examine potential substitutions:

  1. Set u=sin(t)u = \sin(t), which implies du=cos(t)dtdu = \cos(t) dt.
  2. The limits change accordingly: when t=0t = 0, u=sin(0)=0u = \sin(0) = 0, and when t=πt = \pi, u=sin(π)=0u = \sin(\pi) = 0.
  3. The integral becomes more manageable. However, due to the symmetry of sine over the interval, the result will be zero, since the integrand is an odd function (because of the sin(2t)\sin(2t)) over a symmetric interval from 00 to π\pi.

Thus, the value of the integral is:

I=0I = 0

Would you like more details on any of the steps?

Here are 5 questions to consider:

  1. What are the properties of definite integrals that help simplify symmetric intervals?
  2. How can trigonometric identities simplify integrals involving sine and cosine?
  3. What other techniques could be applied to solve integrals involving exponentials and trigonometric functions?
  4. Why does the function sin(2t)\sin(2t) being odd lead to the integral being zero?
  5. How would the integral change if the limits were different?

Tip: Symmetry in integrals is a powerful tool—always check if the integrand is odd or even when dealing with symmetric limits.

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Math Problem Analysis

Mathematical Concepts

Calculus
Definite Integrals
Trigonometric Substitution
Symmetry in Functions

Formulas

cos(t) sin(t) = 1/2 sin(2t)
Integral of an odd function over symmetric limits equals zero

Theorems

Symmetry property of definite integrals
Odd and even function properties

Suitable Grade Level

University Level (Calculus II or higher)