Problem 7 - Active Calculus

Question

For each of the following indefinite integrals, determine whether you would use \(u\) -substitution, integration by parts, neither*, or both to evaluate the integral. In each case, write one sentence to explain your reasoning, and include a statement of any substitutions used. (That is, if you decide in a problem to let \(u=e^{3 x}\), you should state that, as well as that \(\left.d u=3 e^{3 x} d x .\right)\) Finally, use your chosen approach to evaluate each integral. (* one of the following problems does not have an elementary antiderivative and you are not expected to actually evaluate this integral; this will correspond with a choice of "neither" among those given.) a. \(\int x^{2} \cos \left(x^{3}\right) d x\) b. \(\int x^{5} \cos \left(x^{3}\right) d x\left(\right.\) Hint: \(\left.x^{5}=x^{2} \cdot x^{3}\right)\) c. \(\int x \ln \left(x^{2}\right) d x\) d. \(\int \sin \left(x^{4}\right) d x\) e. \(\int x^{3} \sin \left(x^{4}\right) d x\) f. \(\int x^{7} \sin \left(x^{4}\right) d x\)

Step-by-Step Solution

Answer

Part a: \(\frac{1}{3} \sin(x^3) + C\); Part b: \(\frac{1}{3} \left( x^3 \sin(x^3) + \cos(x^3) \right) + C\); Part c: \(\frac{x^2}{2} \ln(x^2) - \frac{x^2}{2} + C\); Part d: Neither; Part e: \(- \frac{1}{4} \cos(x^4) + C\); Part f: \(\frac{1}{4} \left(-x^4 \cos(x^4) + \sin(x^4) \right) + C\).

Step 1: Part (a): Identify the appropriate method

Consider the integral \(\int x^{2} \cos \left(x^{3}\right) d x\). Use \(u\)-substitution because \(x^3\) is inside the cosine function and its derivative (\(3x^2\)) is proportional to \(x^2\).

Step 2: Part (a): Substitution

Let \(u = x^3\). Then \(du = 3x^2 dx\). Adjust the differential: \(dx = \frac{du}{3x^2}\). The integral becomes: \[\int x^{2} \cos \left(x^{3}\right) d x = \int x^2 \cos(u) \frac{du}{3x^2} = \frac{1}{3} \int \cos(u) du.\]

Step 3: Part (a): Integrate

Now integrate: \(\frac{1}{3} \int \cos(u) du = \frac{1}{3} \sin(u) + C\). Substitute back \(u=x^3\): \[\int x^{2} \cos \left(x^{3}\right) d x = \frac{1}{3} \sin(x^3) + C.\]

Step 4: Part (b): Identify the appropriate method

Consider the integral \(\int x^{5} \cos \left(x^{3}\right) d x\). Use \(u\)-substitution because \(x^3\) is inside the cosine function and can be isolated. Also, reformat the polynomial multiplication to emphasize this relationship: \(x^5 = x^2 \cdot x^3\)

Step 5: Part (b): Substitution

Let \(u = x^3\). Then \(du = 3x^2 dx\). Adjust the differential: \(dx = \frac{du}{3x^2}\). Rewrite the integral: \[\int x^5 \cos(x^3) dx = \int x^2 x^3 \cos(x^3) \frac{du}{3x^2} = \int u \cos(u) \frac{du}{3} = \frac{1}{3} \int u \cos(u) du.\]

Step 6: Part (b): Integration by parts

Apply integration by parts: let \(v = u\) and \(dw = \cos(u) du\). Integrate \(dw\) to get \(w = \sin(u)\). The integral becomes \[\frac{1}{3} \left( u \sin(u) - \int \sin(u) du \right) = \frac{1}{3} \left( u \sin(u) + \cos(u) \right).\] Substitute back \(u = x^3\), \[\int x^5 \cos(x^3) dx = \frac{1}{3} \left( x^3 \sin(x^3) + \cos(x^3) \right) + C. \]

Step 7: Part (c): Identify the appropriate method

Consider the integral \(\int x \ln \left(x^{2}\right) d x\). Use integration by parts because we have a product of two functions where one is easily integrable and the other is easily differentiable.

Step 8: Part (c): Integration by parts setup

Let \(u = \ln(x^2)\) and \(dv = x dx\). This means \(du = \frac{2}{x} dx\) and we integrate \(dv\) to get \(v = \frac{x^2}{2} \). Applying integration by parts: \[ \int x \ln(x^2) dx = \frac{x^2}{2} \ln(x^2) - \int \left( \frac{x^2}{2} \cdot \frac{2}{x} \right) dx = \frac{x^2}{2} \ln(x^2) - \int x dx. \]

Step 9: Part (c): Simplifying

This simplifies to: \[ \frac{x^2}{2} \ln(x^2) - \int x dx = \frac{x^2}{2} \ln(x^2) - \frac{x^2}{2} + C. \]

Step 10: Part (d): Identify the appropriate method

Consider the integral \(\int \sin \left(x^{4}\right) d x\). Use 'neither' because there is no elementary antiderivative for \( \sin(x^4) \).

Step 11: Part (e): Identify the appropriate method

Consider the integral \(\int x^3 \sin \left(x^4 \right) dx \). Use \(u\)-substitution because \(x^4\) is inside the sine function and allows for straightforward substitution.

Step 12: Part (e): Substitution

Let \(u = x^4\). Then \(du = 4x^3 dx\). Adjust the differential: \(dx = \frac{du}{4x^3}\). The integral becomes: \[\int x^3 \sin \left(x^4\right) d x = \int x^3 \sin(u) \frac{du}{4x^3} = \frac{1}{4} \int \sin(u) du.\]

Step 13: Part (e): Integrate

Now integrate: \(\frac{1}{4} \int \sin(u) du = -\frac{1}{4} \cos(u) + C\). Substitute back \(u = x^4\): \[\int x^3 \sin \left(x^{4}\right) d x = -\frac{1}{4} \cos(x^4) + C.\]

Step 14: Part (f): Identify the appropriate method

Consider the integral \(\int x^{7} \sin \left(x^{4}\right) d x\). Use \(u\)-substitution first and then integration by parts.'

Step 15: Part (f): Substitution

Let \(u = x^4\). Then \(du = 4x^3 dx\). Adjust the integral: \[\int x^{7} \sin(x^4) dx = \int x^4 x^3 \sin(x^4) dx = \int u x^3 \sin(u) \frac{du}{4x^3} = \frac{1}{4} \int u \sin(u) du. \]

Step 16: Part (f): Integration by parts

Apply integration by parts: let \(v = u\) and \(dw = \sin(u) du\). Integrate \(dw\) to get \(w = -\cos(u)\). The integral becomes \[\frac{1}{4} \left( u (-\cos(u)) - \int -\cos(u) du \right) = \frac{1}{4} \left(-u \cos(u) + \sin(u) \right). \] Substitute back \(u = x^4\), \[\int x^7 \sin(x^4) dx = \frac{1}{4} \left( -x^4 \cos(x^4) + \sin(x^4) \right) + C. \]

Key Concepts

u-substitution

When solving integrals, sometimes a direct approach doesn't work, and that's where substitution methods come in handy. One such method is **u-substitution**. This method involves replacing a complicated expression with a simpler variable, usually denoted as \(u\). The idea is to make the integral easier to evaluate. For example, in the integral \(\ int x^2 \ cos(x^3) \ dx //\), we see that \( x^3 \) is inside a cosine function. The derivative of \( x^3, //\) which is \(3x^2, //\) is proportional to \( x^2 //\). So, we let \( u = x^3 //\). This transforms our integral into a simpler form.

• Step 1: Define the substitution \( u \). For instance, let \( u = x^3 //\).
• Step 2: Compute the differential \( du \). Here, \( du = 3x^2 dx //\).
• Step 3: Adjust the integral by substituting \( u \) and \( du //\). The integral becomes \( \frac{1}{3} \ int cos(u) du \).
• Step 4: Integrate and substitute back. \(\frac{1}{3} \ sin(u) + C //\) returns to \( \frac{1}{3} \ sin(x^3) + C //\).

This method is powerful for integrals involving compositions of functions. Once the substitution turns the integral into a simpler form, standard integration techniques apply.

integration by parts

Another powerful technique for evaluating integrals is **integration by parts**. This method is based on the product rule for differentiation, which is reversed here for integration. It is particularly useful when you have a product of two functions and one of them can be easily differentiated while the other can be easily integrated. The formula for integration by parts is: \( \ int u dv = uv - \ int v du //\).

• Step 1: Identify the parts of the integral. Let \( u //\) be the part to differentiate and \( dv //\) be the part to integrate.
• Step 2: Differentiate \( u //\) to find \( du //\), and integrate \( dv //\) to find \( v //\).
• Step 3: Substitute into \( uv - \ int v du //\).
• Example: For the integral \( \ int x \ ln(x^2) \ dx //\), let \( u = \ ln(x^2) //\) and \( dv = x \ dx //\). Then \( du = \ frac{2}{x} \ dx //\) and \( v = \ frac{x^2}{2} //\). Applying the formula, we get \( \ frac{x^2}{2} \ ln(x^2) - \ int x \ dx //\), which simplifies to \( \ frac{x^2}{2} \ ln(x^2) - \ frac{x^2}{2} + C //\).

By carefully choosing \( u //\) and \( dv //\), this method simplifies the integral calculation significantly.

integral evaluation

Once you have chosen the appropriate method to simplify the integral, the final step is the **integral evaluation**. This involves performing actual integration and simplifying the results. Correct evaluation ensures you have the necessary antiderivatives and constants in place.

• For substitution or u-substitution, integrate the simplified form and revert to original variables.
• For integration by parts, carefully handle the resulting terms and any leftover integrals.
• Apply constants of integration (\( +C //\)) is crucial for indefinite integrals.
• Example: In \( \ int x^3 \ sin(x^4) \ dx //\), letting \( u = x^4 //\) results in \( \ frac{1}{4} \ int \ sin(u) du //\). Integrating gives \( -\frac{1}{4} \ cos(u) + C //\), substituting back yields \( -\frac{1}{4} \ cos(x^4) + C //\).

Always double-check your work by differentiating your result to see if you retrieve the original integrand. This not only ensures correctness but also deepens understanding of the integral concepts.