You appear to be asking for an integral involving the given terms. Let me interpret the integral as follows:

$\int_{\pi/4} \sin(t) \cos^2(t) \, dt$

To solve this:

Step 1: Simplify the integral

Using the trigonometric identity: $\cos^2(t) = 1 - \sin^2(t)$ The integral becomes: $\int_{\pi/4} \sin(t) (1 - \sin^2(t)) \, dt$

Step 2: Use substitution

Let: $u = \sin(t) \quad \text{so that} \quad du = \cos(t) \, dt$

Change the limits of integration for $u$:

• When $t = \pi/4$, $u = \sin(\pi/4) = \frac{\sqrt{2}}{2}$.

Thus, the integral becomes: $\int \sin(t) (1 - \sin^2(t)) \, dt = \int_{\sqrt{2}/2}^{u} u(1 - u^2) \, du$

Step 3: Expand and integrate

Expand $u(1 - u^2)$: $\int_{\sqrt{2}/2} u(1 - u^2) \, du = \int_{\sqrt{2}/2} (u - u^3) \, du$

Now integrate each term: $\int u \, du = \frac{u^2}{2}, \quad \int u^3 \, du = \frac{u^4}{4}$

Thus: $\int_{\sqrt{2}/2} u(1 - u^2) \, du = \left[\frac{u^2}{2} - \frac{u^4}{4} \right]_{\sqrt{2}/2}^1$

Step 4: Evaluate the limits

First, calculate at $u = 1$: $\frac{1^2}{2} - \frac{1^4}{4} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$

Now, calculate at $u = \frac{\sqrt{2}}{2}$: $\frac{\left(\frac{\sqrt{2}}{2}\right)^2}{2} - \frac{\left(\frac{\sqrt{2}}{2}\right)^4}{4} = \frac{\frac{2}{4}}{2} - \frac{\frac{4}{16}}{4} = \frac{1}{4} - \frac{1}{16} = \frac{4}{16} - \frac{1}{16} = \frac{3}{16}$

Subtract the results: $\frac{1}{4} - \frac{3}{16} = \frac{4}{16} - \frac{3}{16} = \frac{1}{16}$

Final Answer:

$\int_{\pi/4} \sin(t) \cos^2(t) \, dt = \frac{1}{16}$

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Do you need further details or clarifications?

Expanding Questions:

1. Would you like to verify each substitution step?
2. Do you want a graph illustrating this integral geometrically?
3. Shall I explain the trigonometric identity $\cos^2(t) = 1 - \sin^2(t)$ in detail?
4. Would you like to explore alternative methods to solve this integral?
5. Are you interested in practical applications of such integrals?

Tip:

Always double-check the limits of integration after substitution to avoid errors in definite integrals!