The integral you provided is:

$\int_0^t u \cos^2(u) \, du.$

To solve this, we can apply a strategy to simplify the integrand. First, recall the trigonometric identity:

$\cos^2(u) = \frac{1 + \cos(2u)}{2}.$

Using this identity, the integral becomes:

$\int_0^t u \cos^2(u) \, du = \int_0^t u \cdot \frac{1 + \cos(2u)}{2} \, du = \frac{1}{2} \int_0^t u (1 + \cos(2u)) \, du.$

We now split the integral:

$\frac{1}{2} \int_0^t u (1 + \cos(2u)) \, du = \frac{1}{2} \left( \int_0^t u \, du + \int_0^t u \cos(2u) \, du \right).$

First Integral:

$\int_0^t u \, du = \frac{u^2}{2} \Big|_0^t = \frac{t^2}{2}.$

Second Integral:

For $\int_0^t u \cos(2u) \, du$, we can use integration by parts. Let:

• $v = u$, so $dv = du$,
• $dw = \cos(2u) \, du$, so $w = \frac{\sin(2u)}{2}$.

Thus, using the integration by parts formula $\int v \, dw = v w - \int w \, dv$:

$\int u \cos(2u) \, du = \frac{u \sin(2u)}{2} - \frac{1}{2} \int \sin(2u) \, du.$

Now, the remaining integral is:

$\int \sin(2u) \, du = -\frac{\cos(2u)}{2}.$

So, the full expression becomes:

$\int u \cos(2u) \, du = \frac{u \sin(2u)}{2} + \frac{\cos(2u)}{4}.$

Evaluating from $0$ to $t$, we get:

$\int_0^t u \cos(2u) \, du = \frac{t \sin(2t)}{2} + \frac{\cos(2t)}{4} - \left( 0 + \frac{1}{4} \right) = \frac{t \sin(2t)}{2} + \frac{\cos(2t)}{4} - \frac{1}{4}.$

Final Result:

Now, summing both parts:

$\frac{1}{2} \left( \frac{t^2}{2} + \frac{t \sin(2t)}{2} + \frac{\cos(2t)}{4} - \frac{1}{4} \right).$

Simplifying this expression:

$\frac{t^2}{4} + \frac{t \sin(2t)}{4} + \frac{\cos(2t)}{8} - \frac{1}{8}.$

Thus, the value of the integral is:

$\int_0^t u \cos^2(u) \, du = \frac{t^2}{4} + \frac{t \sin(2t)}{4} + \frac{\cos(2t)}{8} - \frac{1}{8}.$

Would you like further details or have any questions? Here are some related questions for practice:

1. How does the trigonometric identity for $\cos^2(u)$ simplify integrals?
2. Can you explain the steps involved in applying integration by parts to this problem?
3. What are the general techniques for solving integrals involving products of functions?
4. How would the result change if the limits of integration were different?
5. What are the applications of integrals involving trigonometric functions in physics or engineering?

Tip: Always check if a trigonometric identity can simplify the integrand before proceeding with more advanced techniques like integration by parts.