Integration By Parts Examples With Solutions

Integration by Parts: Examples with Solutions

Integration by parts is a powerful technique used in calculus to solve integrals that cannot be easily solved using standard integration rules. It's particularly useful when dealing with integrals involving products of functions. This comprehensive guide will walk you through the core concept, provide numerous examples with detailed solutions, and offer tips and tricks to master this essential integration method.

Understanding Integration by Parts

The formula for integration by parts stems directly from the product rule for differentiation. Recall that the product rule states:

d(uv)/dx = u(dv/dx) + v(du/dx)

Rearranging this equation and integrating both sides with respect to x, we get the integration by parts formula:

∫u(dv/dx)dx = uv - ∫v(du/dx)dx

This can be more concisely written as:

∫u dv = uv - ∫v du

The key to successfully applying integration by parts lies in choosing the appropriate 'u' and 'dv'. A helpful mnemonic device is LIATE, which prioritizes the order of function types:

• Logarithmic functions
• Inverse trigonometric functions
• Algebraic functions (polynomials)
• Trigonometric functions
• Exponential functions

Generally, you should choose 'u' to be the function that simplifies when differentiated, and 'dv' to be the function that is easily integrated. Let's explore this through various examples.

Example 1: ∫x cos(x) dx

Here, we choose:

• u = x => du = dx
• dv = cos(x) dx => v = sin(x)

Applying the integration by parts formula:

∫x cos(x) dx = x sin(x) - ∫sin(x) dx = x sin(x) + cos(x) + C

Where 'C' represents the constant of integration.

Solution: x sin(x) + cos(x) + C

Example 2: ∫x² eˣ dx

For this integral, we'll need to apply integration by parts twice.

First Application:

• u = x² => du = 2x dx
• dv = eˣ dx => v = eˣ

∫x² eˣ dx = x²eˣ - ∫2x eˣ dx

Second Application:

Now we need to solve ∫2x eˣ dx. Let's apply integration by parts again:

• u = 2x => du = 2 dx
• dv = eˣ dx => v = eˣ

∫2x eˣ dx = 2x eˣ - ∫2 eˣ dx = 2x eˣ - 2eˣ

Substituting this back into our original equation:

∫x² eˣ dx = x²eˣ - (2x eˣ - 2eˣ) = x²eˣ - 2x eˣ + 2eˣ + C

Solution: x²eˣ - 2x eˣ + 2eˣ + C

Example 3: ∫ln(x) dx

This example showcases integration by parts with a logarithmic function. A common trick here is to let 'u' be the logarithmic function, even though it isn't in the LIATE order. This is because the derivative of ln(x) is simpler.

• u = ln(x) => du = (1/x) dx
• dv = dx => v = x

∫ln(x) dx = x ln(x) - ∫x (1/x) dx = x ln(x) - ∫dx = x ln(x) - x + C

Solution: x ln(x) - x + C

Example 4: ∫x³ ln(x) dx

This combines algebraic and logarithmic functions. Following LIATE, we choose:

• u = ln(x) => du = (1/x) dx
• dv = x³ dx => v = (x⁴)/4

∫x³ ln(x) dx = (x⁴/4) ln(x) - ∫(x⁴/4)(1/x) dx = (x⁴/4) ln(x) - ∫(x³/4) dx

= (x⁴/4) ln(x) - (x⁴/16) + C

Solution: (x⁴/4) ln(x) - (x⁴/16) + C

Example 5: ∫eˣ sin(x) dx

This integral requires a clever trick. We'll apply integration by parts twice and then solve for the original integral.

First Application:

• u = sin(x) => du = cos(x) dx
• dv = eˣ dx => v = eˣ

∫eˣ sin(x) dx = eˣ sin(x) - ∫eˣ cos(x) dx

Second Application:

Now we need to solve ∫eˣ cos(x) dx:

• u = cos(x) => du = -sin(x) dx
• dv = eˣ dx => v = eˣ

∫eˣ cos(x) dx = eˣ cos(x) + ∫eˣ sin(x) dx

Substituting this back into our original equation:

∫eˣ sin(x) dx = eˣ sin(x) - [eˣ cos(x) + ∫eˣ sin(x) dx]

Notice that the original integral appears on both sides of the equation. Let's call it 'I':

I = eˣ sin(x) - eˣ cos(x) - I

2I = eˣ sin(x) - eˣ cos(x)

I = (eˣ sin(x) - eˣ cos(x))/2 + C

Solution: (eˣ sin(x) - eˣ cos(x))/2 + C

Example 6: ∫arctan(x) dx

This involves an inverse trigonometric function.

• u = arctan(x) => du = 1/(1+x²) dx
• dv = dx => v = x

∫arctan(x) dx = x arctan(x) - ∫x/(1+x²) dx

The integral ∫x/(1+x²) dx can be solved using substitution: Let w = 1 + x², then dw = 2x dx.

∫x/(1+x²) dx = (1/2) ∫dw/w = (1/2) ln|w| = (1/2) ln|1 + x²|

Therefore:

∫arctan(x) dx = x arctan(x) - (1/2) ln|1 + x²| + C

Solution: x arctan(x) - (1/2) ln|1 + x²| + C

Advanced Considerations and Tips

• Repeated Integration by Parts: As seen in examples 2 and 5, sometimes you need to apply integration by parts multiple times to reach a solvable integral.

• Tabular Integration: For integrals involving repeated applications of integration by parts, especially with polynomials multiplied by exponential or trigonometric functions, tabular integration provides a systematic and efficient approach.

• Choosing 'u' and 'dv': The choice of 'u' and 'dv' is crucial. While LIATE provides a guideline, sometimes experimentation is needed to find the most effective approach.

• Recognizing Patterns: Look for patterns and recurring integrals within your calculations. This often points towards a solution or a need for a different integration strategy.

• Checking Your Work: Always differentiate your solution to verify that you get the original integrand. This helps catch errors.

Conclusion

Integration by parts is a fundamental technique in calculus. Mastering it requires practice and a good understanding of the underlying principles. By carefully selecting 'u' and 'dv', applying the formula correctly, and persistently practicing, you will become proficient in solving a wide range of complex integrals. Remember to always check your solutions by differentiation to ensure accuracy. The examples provided, along with the tips and strategies discussed, will equip you with the necessary tools and understanding to tackle a diverse range of integration by parts problems with confidence. Remember to utilize the strategies mentioned above to maximize the effectiveness of your problem-solving approach. Consistent practice will solidify your understanding and refine your ability to quickly identify the optimal solution path.