91Ó°ÊÓ

An indefinite integral is essentially the reverse operation of differentiation. When we talk about finding the indefinite integral of a function, we aim to find a general formula that describes antiderivatives of the function. This means we're looking for all possible functions which, if you differentiate them, would give back the original function we started with. In this exercise, we're focusing on the indefinite integral of the function:\[f(x)=\sin \left(\frac{x}{3}\right)+\cos \left(\frac{x}{3}\right)\]Finding the indefinite integral means determining the function \(F(x)\) such that \(F'(x) = f(x)\). The result will include a constant of integration \(C\), reflecting that there are infinitely many antiderivatives for a given function. In our solution, this means looking for \(F(x)\) which is the general form:\[F(x)= -3 \cos \left(\frac{x}{3}\right) + 3 \sin \left(\frac{x}{3}\right) + C\]

Trigonometric functions like sine and cosine are periodic functions that are fundamental in many areas of mathematics, including calculus. When you take the indefinite integral of a trigonometric function, you find another trigonometric function.

• The integral of \(\sin(x)\) is \(-\cos(x)\).
• The integral of \(\cos(x)\) is \(\sin(x)\).

In the given exercise, the function \(\sin \left(\frac{x}{3}\right)+\cos \left(\frac{x}{3}\right)\) involves both \(\sin\) and \(\cos\) functions. Understanding these intefrals helps calculate their indefinite integrals effectively:

• For \(\sin\left(\frac{x}{3}\right)\), you move towards \(-\cos\left(\frac{x}{3}\right)\), adjusting by constants from the substitution (covered next).
• For \(\cos\left(\frac{x}{3}\right)\), you arrive at \(\sin\left(\frac{x}{3}\right)\), again adjusting similarly with the substitution method.

The substitution method is a crucial technique in calculus to simplify integration of complex functions. This approach often makes the integration process much more manageable by transforming it into a form that's easier to handle. In this problem, we encounter a situation where direct integration is complicated, and substitution is ideal.

Setting the Substitution

For the function \(f(x)\), substituting \(u = \frac{x}{3}\) simplifies our integrations. The differential \(dx\) becomes a function of \(du\), specifically, \(dx = 3 \, du\).

Applying to the Function

When applying this substitution:

• The \(\sin\left(\frac{x}{3}\right)\) term becomes \(-3 \cos(u)\) as \(\int \sin(u) \, du = -\cos(u)\).
• The \(\cos\left(\frac{x}{3}\right)\) term becomes \(3 \sin(u)\) as \(\int \cos(u) \, du = \sin(u)\).

Reversing the Substitution

After finding the antiderivatives in terms of \(u\), revert back to \(x\) by replacing \(u\) with \(\frac{x}{3}\), giving us the integrated function. Combing these results and adding the constant of integration \(C\), we acquire the general antiderivative:\[F(x) = -3 \cos\left(\frac{x}{3}\right) + 3 \sin\left(\frac{x}{3}\right) + C\]