Math Problem Statement

Evaluate the integral: \( \int \frac{125e^{3t}}{\sqrt{9 + 25e^{2t}}} dt \)

Solution

The given integral is:

125e3t9+25e2tdt\int \frac{125 e^{3t}}{\sqrt{9 + 25 e^{2t}}} \, dt

Solution:

To solve this, we can make a substitution to simplify the integrand. Let's define u=etu = e^t, so that du=etdtdu = e^t dt. This also implies that e2t=u2e^{2t} = u^2 and e3t=u3e^{3t} = u^3.

Rewriting the integral in terms of uu:

125u39+25u2duu\int \frac{125 u^3}{\sqrt{9 + 25 u^2}} \cdot \frac{du}{u}

This simplifies to:

125u29+25u2du\int \frac{125 u^2}{\sqrt{9 + 25 u^2}} \, du

Now, factor out constants from the square root:

9+25u2=9(1+259u2)=31+259u2\sqrt{9 + 25 u^2} = \sqrt{9(1 + \frac{25}{9}u^2)} = 3\sqrt{1 + \frac{25}{9} u^2}

Thus, the integral becomes:

125u231+259u2du\int \frac{125 u^2}{3\sqrt{1 + \frac{25}{9} u^2}} \, du

Now, to simplify this further, let’s make another substitution. Let v=53uv = \frac{5}{3} u, so that dv=53dudv = \frac{5}{3} du, and the integral becomes:

125(35v)231+v235dv\int \frac{125 \cdot \left(\frac{3}{5}v\right)^2}{3\sqrt{1 + v^2}} \cdot \frac{3}{5} dv

This further simplifies to a standard form that can be evaluated using trigonometric or hyperbolic identities.


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Math Problem Analysis

Mathematical Concepts

Calculus
Integration
Substitution Method
Trigonometric Substitution

Formulas

Substitution: u = e^t, du = e^t dt
Square root simplification: \( \sqrt{9 + 25u^2} = 3 \sqrt{1 + \frac{25}{9} u^2} \)
Trigonometric substitution: \( v = \frac{5}{3} u \)

Theorems

Basic integration techniques
Trigonometric substitution
Chain rule for derivatives

Suitable Grade Level

Undergraduate Calculus