Let me analyze the image and extract the mathematical problem for solving.The problem is to evaluate the integral:

$\int_{\pi/6}^{\pi/2} \frac{\cos^3 x}{\sqrt{\sin x}} \, dx = \frac{a}{5} - \frac{b}{10\sqrt{2}}, \quad \text{where } a = \, ? \, \text{and } b = \, ?.$

We aim to solve the integral step by step and identify the values of $a$ and $b$.

Step 1: Simplify the integrand

Rewrite $\cos^3 x$ using $\cos^3 x = \cos x \cdot \cos^2 x$, and use the trigonometric identity $\cos^2 x = 1 - \sin^2 x$: $\frac{\cos^3 x}{\sqrt{\sin x}} = \frac{\cos x (1 - \sin^2 x)}{\sqrt{\sin x}}.$

This becomes: $\frac{\cos x}{\sqrt{\sin x}} - \frac{\cos x \sin^2 x}{\sqrt{\sin x}}.$

Simplify each term: $\frac{\cos x}{\sqrt{\sin x}} - \sqrt{\sin x} \cos x.$

Step 2: Substitution

Let $u = \sin x$, so that $du = \cos x \, dx$. The limits of integration transform as follows:

• When $x = \pi/6$, $u = \sin(\pi/6) = 1/2$.
• When $x = \pi/2$, $u = \sin(\pi/2) = 1$.

The integral becomes: $\int_{1/2}^{1} \frac{1}{\sqrt{u}} \, du - \int_{1/2}^{1} \sqrt{u} \, du.$

Step 3: Solve each integral

1. For $\int \frac{1}{\sqrt{u}} \, du$: $\int \frac{1}{\sqrt{u}} \, du = 2\sqrt{u}.$ Evaluating from $u = 1/2$ to $u = 1$: $2\sqrt{1} - 2\sqrt{\frac{1}{2}} = 2 - 2\cdot\frac{1}{\sqrt{2}} = 2 - \sqrt{2}.$

2. For $\int \sqrt{u} \, du$: $\int \sqrt{u} \, du = \frac{2}{3} u^{3/2}.$ Evaluating from $u = 1/2$ to $u = 1$: $\frac{2}{3}(1^{3/2}) - \frac{2}{3}\left(\frac{1}{2}\right)^{3/2} = \frac{2}{3} - \frac{2}{3} \cdot \frac{1}{2\sqrt{2}} = \frac{2}{3} - \frac{1}{3\sqrt{2}}.$

Step 4: Combine results

The integral becomes: $\left(2 - \sqrt{2}\right) - \left(\frac{2}{3} - \frac{1}{3\sqrt{2}}\right).$

Simplify: $2 - \sqrt{2} - \frac{2}{3} + \frac{1}{3\sqrt{2}}.$

Combine terms: $\left(2 - \frac{2}{3}\right) - \left(\sqrt{2} - \frac{1}{3\sqrt{2}}\right) = \frac{6}{3} - \frac{2}{3} - \sqrt{2} + \frac{1}{3\sqrt{2}} = \frac{4}{3} - \sqrt{2} + \frac{1}{3\sqrt{2}}.$

Step 5: Match the form

The result is in the form: $\frac{a}{5} - \frac{b}{10\sqrt{2}}.$

Equating coefficients, we find: $\frac{a}{5} = \frac{4}{3} \quad \Rightarrow \quad a = \frac{20}{3},$ $\frac{b}{10\sqrt{2}} = \sqrt{2} - \frac{1}{3\sqrt{2}} \quad \Rightarrow \quad b = 10\left(2 - \frac{1}{3}\right) = \frac{50}{3}.$

Final Answer:

$a = \frac{20}{3}, \quad b = \frac{50}{3}.$

Let me know if you'd like further details! Here are 5 related questions for practice:

1. How do trigonometric identities help simplify integrals involving $\cos^n x$?
2. What is the purpose of substitution in solving integrals?
3. How do you evaluate definite integrals after substitution?
4. Why is it helpful to express results in specific forms (e.g., $\frac{a}{5} - \frac{b}{10\sqrt{2}}$)?
5. What is the geometric interpretation of this integral?

Tip: Always check your substitution limits carefully when performing definite integrals!