https://file.aiquickdraw.com/mathbot/file/1734292086238e3wbo074.jpg

The integral in the image is written as:

\[
\int_{\pi/4}^{\pi/2} \int_{0}^{\frac{4}{\sin(\theta)}} r^2 \cos(\theta) \, dr \, d\theta
\]

This is a double integral in polar coordinates. Let me provide an explanation of its components and how to solve it step-by-step:

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### Step 1: Understanding the limits and the integrand
- The **inner integral** \(\int_0^{\frac{4}{\sin(\theta)}}\) integrates with respect to \(r\), where the upper limit depends on \(\theta\).
- The **outer integral** \(\int_{\pi/4}^{\pi/2}\) integrates with respect to \(\theta\).
- The integrand is \(r^2 \cos(\theta)\), where \(r\) represents the radial coordinate and \(\theta\) is the angular coordinate.

---

### Step 2: Solving the inner integral (with respect to \(r\))
The inner integral is:

\[
\int_0^{\frac{4}{\sin(\theta)}} r^2 \cos(\theta) \, dr
\]

Since \(\cos(\theta)\) is constant with respect to \(r\), it can be factored out:

\[
\int_0^{\frac{4}{\sin(\theta)}} r^2 \cos(\theta) \, dr = \cos(\theta) \int_0^{\frac{4}{\sin(\theta)}} r^2 \, dr
\]

The integral of \(r^2\) is:

\[
\int r^2 \, dr = \frac{r^3}{3}
\]

Now evaluate from \(0\) to \( \frac{4}{\sin(\theta)} \):

\[
\int_0^{\frac{4}{\sin(\theta)}} r^2 \, dr = \frac{\left( \frac{4}{\sin(\theta)} \right)^3}{3} - 0 = \frac{4^3}{3 \sin^3(\theta)} = \frac{64}{3 \sin^3(\theta)}
\]

Thus, the inner integral becomes:

\[
\int_0^{\frac{4}{\sin(\theta)}} r^2 \cos(\theta) \, dr = \cos(\theta) \cdot \frac{64}{3 \sin^3(\theta)}
\]

Simplify:

\[
\int_0^{\frac{4}{\sin(\theta)}} r^2 \cos(\theta) \, dr = \frac{64 \cos(\theta)}{3 \sin^3(\theta)}
\]

---

### Step 3: Solving the outer integral (with respect to \(\theta\))
The outer integral is now:

\[
\int_{\pi/4}^{\pi/2} \frac{64 \cos(\theta)}{3 \sin^3(\theta)} \, d\theta
\]

Factor out the constant \(\frac{64}{3}\):

\[
\frac{64}{3} \int_{\pi/4}^{\pi/2} \frac{\cos(\theta)}{\sin^3(\theta)} \, d\theta
\]

To solve this integral, recognize that:

\[
\frac{\cos(\theta)}{\sin^3(\theta)} = \frac{d}{d\theta} \left( -\frac{1}{2 \sin^2(\theta)} \right)
\]

---

### Step 4: Substitution and evaluation
The derivative of \(-\frac{1}{2 \sin^2(\theta)}\) simplifies the integral:

\[
\int \frac{\cos(\theta)}{\sin^3(\theta)} \, d\theta = -\frac{1}{2 \sin^2(\theta)}
\]

Evaluate this from \(\theta = \pi/4\) to \(\theta = \pi/2\):

1. At \(\theta = \pi/2\), \(\sin(\pi/2) = 1\):

\[
-\frac{1}{2 \sin^2(\pi/2)} = -\frac{1}{2 \cdot 1^2} = -\frac{1}{2}
\]

2. At \(\theta = \pi/4\), \(\sin(\pi/4) = \frac{\sqrt{2}}{2}\):

\[
-\frac{1}{2 \sin^2(\pi/4)} = -\frac{1}{2 \left( \frac{\sqrt{2}}{2} \right)^2} = -\frac{1}{2 \cdot \frac{1}{2}} = -1
\]

---

### Step 5: Combine results
Subtract the values:

\[
\left[ -\frac{1}{2} \right] - \left[ -1 \right] = -\frac{1}{2} + 1 = \frac{1}{2}
\]

Multiply by the constant \(\frac{64}{3}\):

\[
\frac{64}{3} \cdot \frac{1}{2} = \frac{64}{6} = \frac{32}{3}
\]

---

### Final Answer:
The value of the double integral is:

\[
\frac{32}{3}
\]

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### Let me know if you need further clarifications! Here are 5 related questions:
1. How do you convert rectangular integrals to polar form?
2. What is the geometric interpretation of double integrals in polar coordinates?
3. How do you integrate functions involving \(\sin(\theta)\) and \(\cos(\theta)\)?
4. How do you evaluate improper integrals in polar form?
5. What are the limits of integration for polar regions?

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**Tip:** Always look for symmetry in polar integrals; it often simplifies calculations!

Convert the given rectangular integral into an equivalent polar integral and solve:
∫ from π/4 to π/2 ∫ from 0 to 4/sin(θ) r² cos(θ) dr dθ

Learn to convert and solve the double integral ∫ from π/4 to π/2 ∫ from 0 to 4/sin(θ) r² cos(θ) dr dθ using polar coordinates. This step-by-step guide covers key integration techniques, trigonometric simplifications, and final solutions.

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