Integral Calculation Exercise 3

We aim to evaluate the following integral:

$$ \int x \cdot \cos x \ dx $$

This can be efficiently solved using integration by parts, where we set \( f(x) = x \) and \( g'(x) = \cos x \).

The integration by parts formula is:

$$ \int f(x) \cdot g'(x) \, dx = f(x) \cdot g(x) - \int f'(x) \cdot g(x) \, dx $$

Since the antiderivative of \( \cos x \) is \( \sin x \), we have:

$$ \int x \cdot \cos x \, dx = x \cdot \sin x - \int f'(x) \cdot \sin x \, dx $$

Now, \( f'(x) = 1 \), so the expression simplifies to:

$$ \int x \cdot \cos x \, dx = x \cdot \sin x - \int \sin x \, dx $$

This remaining integral is straightforward to compute.

The antiderivative of \( \sin x \) is \( -\cos x \), so:

$$ \int x \cdot \cos x \, dx = x \cdot \sin x - ( - \cos x ) + c $$

Therefore, the final result is:

$$ \int x \cdot \cos x \, dx = x \cdot \sin x + \cos x + c $$

And so on.