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The image asks to evaluate the integral:

\[
\int_1^3 x \sin^4(x) \, dx
\]

Let's solve this step by step.

### Step 1: Simplify the Integral Using Trigonometric Identity
The integral involves \(\sin^4(x)\). We can use the trigonometric identity:

\[
\sin^4(x) = \left(\frac{1 - \cos(2x)}{2}\right)^2
\]

Expanding this identity gives:

\[
\sin^4(x) = \frac{1}{4} \left(1 - 2\cos(2x) + \cos^2(2x)\right)
\]

### Step 2: Expand the Integral
Substitute this expression into the integral:

\[
\int_1^3 x \sin^4(x) \, dx = \frac{1}{4} \int_1^3 x \left(1 - 2\cos(2x) + \cos^2(2x)\right) dx
\]

This can be expanded as:

\[
\frac{1}{4} \left[ \int_1^3 x \, dx - 2\int_1^3 x\cos(2x) \, dx + \int_1^3 x\cos^2(2x) \, dx \right]
\]

### Step 3: Solve Each Integral Separately
1. **First Integral: \(\int_1^3 x \, dx\)**

   This is straightforward:

   \[
   \int_1^3 x \, dx = \left. \frac{x^2}{2} \right|_1^3 = \frac{9}{2} - \frac{1}{2} = 4
   \]

2. **Second Integral: \(\int_1^3 x \cos(2x) \, dx\)**

   Use integration by parts. Let \( u = x \) and \( dv = \cos(2x) \, dx \).

   \[
   du = dx, \quad v = \frac{1}{2} \sin(2x)
   \]

   Therefore,

   \[
   \int x \cos(2x) \, dx = \frac{1}{2} x \sin(2x) - \frac{1}{2} \int \sin(2x) \, dx
   \]

   The integral \( \int \sin(2x) \, dx \) is:

   \[
   \int \sin(2x) \, dx = -\frac{1}{2} \cos(2x)
   \]

   Substituting back,

   \[
   \int x \cos(2x) \, dx = \frac{1}{2} x \sin(2x) + \frac{1}{4} \cos(2x)
   \]

   Evaluate this from 1 to 3:

  

Evaluate the integral ∫ from 1 to 3 of x*sin^4(x) dx.

Learn how to evaluate the integral of x*sin^4(x) from 1 to 3 using trigonometric identities and integration by parts. This step-by-step solution is suitable for Calculus students at the undergraduate level.

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