Problem 60 - Calculus Single Variable

Question

Make an appropriate substitution before integrating by parts. $$ \int 3 x^{5} \sin \left(x^{3}\right) d x $$

Step-by-Step Solution

Answer

The integral is \(-x^3 \cos(x^3) + \sin(x^3) + C\).

Step 1: Identify the substitution

To make the integration easier, we'll first make a substitution. Notice that inside the sine function, we have \(x^3\). Let's set \(u = x^3\). This substitution will simplify the expression inside the sine function.

Step 2: Compute the differential

Differentiate \(u = x^3\) to find \(du\). This gives \(du = 3x^2 \, dx\). Solve for \(dx\) to get \(dx = \frac{du}{3x^2}\).

Step 3: Rewrite the integral using substitution

Substitute \(u = x^3\) and \(dx\) into the integral. Then the integral becomes: \(\int 3 x^5 \sin (u) \frac{du}{3x^2}\). Simplifying this, we get \(\int x^3 \sin(u) \, du\). Given \(u = x^3\), this means simply \(\int u \sin(u) \, du\).

Step 4: Integration by Parts

Use integration by parts for \(\int u \sin(u) \, du\). Let \(v = u\), so \(dv = du\), and let \(w = \sin(u)\), so \(dw = \cos(u) \, du\). The integration by parts formula \(\int v \, dw = vw - \int w \, dv\) gives: \(u(-\cos(u)) - \int -\cos(u) \, du\).

Step 5: Complete the integration

Calculate \(\int \cos(u) \, du\) which is \(\sin(u)\). This results in the expression: \(-u \cos(u) + \sin(u) + C\).

Step 6: Substitute back in terms of x

Now revert \(u\) back to \(x^3\) to express the integral in terms of \(x\). Substitute \(u = x^3\) back, giving \(-x^3 \cos(x^3) + \sin(x^3) + C\).

Key Concepts

Substitution Method

The substitution method, also known as u-substitution, is a powerful technique in integration, allowing us to simplify complex integrals. It works by introducing a new variable that simplifies part of the integrand (the function being integrated).

In this exercise, we identified that the expression within the sine function, specifically \(x^{3}\), could make the entire integration simpler if viewed as a separate variable \(u\). By setting \(u = x^3\), we can transform the complex expression into something more manageable by focusing only on \(u\).

• Substitution helps in reducing the complexity of the original integrand.
• It requires computing the derivative of the substitution, which in our case was \(du = 3x^2 \, dx\).
• The idea is to replace all parts of the integral related to \(x\) with expressions in terms of \(u\).

This change of variables makes it easier to apply further techniques, such as integration by parts. Remember, the goal is to transform difficult integrals into ones we can solve easily.

Integration by Parts

Integration by parts is another fundamental integration technique, which is useful when dealing with products of functions. The formula is derived from the product rule for differentiation:
\[ \int v \, dw = vw - \int w \, dv \]

In our exercise, once substitution reduced the integral to \(\int u \sin(u) \, du\), it became evident that using integration by parts would be the next step.

• By choosing \(v = u\) and \(dw = \sin(u) \, du\), the derivatives required, \(dv = du\) and \(w = -\cos(u)\), were straightforward to determine.
• This led to a simplified evaluation: \( -u \cos(u) + \int \cos(u) \, du\).

Applying integration by parts is often straightforward but choosing the right \(v\) and \(dw\) can sometimes require a bit of practice.

It's essential to look at the integral components and decide which parts will continue to simplify the process. This method steps in when simple u-substitutions aren’t enough.

Differential Calculus

Differential calculus is the mathematical study of how functions change. It involves derivatives, which provide the rate of change of one quantity with respect to another.

In our exercise, identifying the differential \(du = 3x^2 \, dx\) was a vital part of using the substitution method. Differential calculus allowed us to translate between differing variable perspectives, from \(x\) to \(u\), ensuring we properly accounted for how each infinitesimal change in \(x\) affected \(u\).

• The concept of differentiating \(x^3\) to get \(3x^2\), provided the necessary toolset for the back and forth between his variables.
• Every substitution or change of variables hinges on understanding these foundational differentiation rules.

Without differential calculus, approaches like substitution would not be possible, as they rely inherently on capacity to interpret and manipulate derivatives. This makes differential calculus an indispensable part of tackling integrals.