Integration By Parts Examples With Solutions

Integration by parts is a powerful technique in calculus used to evaluate integrals, especially when the integrand is a product of two functions. It's derived from the product rule for differentiation and provides a way to "reverse" the product rule in integration. Mastering integration by parts unlocks the ability to solve a wide variety of complex integrals, making it an indispensable tool for students and professionals in mathematics, physics, engineering, and related fields. This comprehensive guide provides a detailed explanation of integration by parts, accompanied by numerous examples with step-by-step solutions to help you grasp the concept and apply it effectively.

Understanding Integration by Parts

The integration by parts formula is derived directly from the product rule of differentiation. Recall that the product rule states:

d/dx [u(x)v(x)] = u'(x)v(x) + u(x)v'(x)

Integrating both sides with respect to x, we get:

∫ d/dx [u(x)v(x)] dx = ∫ u'(x)v(x) dx + ∫ u(x)v'(x) dx

u(x)v(x) = ∫ v(x)u'(x) dx + ∫ u(x)v'(x) dx

Rearranging this equation yields the integration by parts formula:

∫ u dv = uv - ∫ v du

Where:

• u is a function of x that we choose to differentiate.
• dv is the remaining part of the integrand, which we choose to integrate.
• du is the derivative of u with respect to x.
• v is the integral of dv with respect to x.

The key to successfully applying integration by parts lies in choosing appropriate functions for u and dv. The goal is to select u and dv such that the integral ∫ v du is simpler than the original integral ∫ u dv. A common mnemonic to aid in this selection is LIATE or ILATE, which prioritizes function types for u in the following order:

• L: Logarithmic functions (e.g., ln(x), log₂(x))
• I: Inverse trigonometric functions (e.g., arctan(x), arcsin(x))
• A: Algebraic functions (e.g., x², x³ + 1)
• T: Trigonometric functions (e.g., sin(x), cos(x))
• E: Exponential functions (e.g., eˣ, 2ˣ)

While LIATE/ILATE provides a helpful guideline, it's not a rigid rule. Sometimes, choosing u and dv based on other considerations, such as simplifying the resulting integral, might be more effective.

Integration by Parts: Examples with Solutions

Let's illustrate the application of integration by parts through several examples with detailed solutions.

Example 1: ∫ x cos(x) dx

1. Choose u and dv:

   Following LIATE, we have:

   • Algebraic function: x
   • Trigonometric function: cos(x)

   Therefore, we choose:

   • u = x
   • dv = cos(x) dx

2. Find du and v:

   • du = dx (derivative of u)
   • v = ∫ cos(x) dx = sin(x) (integral of dv)

3. Apply the Integration by Parts Formula:

   ∫ u dv = uv - ∫ v du

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

4. Evaluate the Remaining Integral:

   ∫ sin(x) dx = -cos(x)

5. Substitute and Add the Constant of Integration:

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

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

Example 2: ∫ x eˣ dx

1. Choose u and dv:

   Following LIATE:

   • Algebraic function: x
   • Exponential function: eˣ

   Therefore, we choose:

   • u = x
   • dv = eˣ dx

2. Find du and v:

   • du = dx
   • v = ∫ eˣ dx = eˣ

3. Apply the Integration by Parts Formula:

   ∫ u dv = uv - ∫ v du

   ∫ x eˣ dx = x eˣ - ∫ eˣ dx

4. Evaluate the Remaining Integral:

   ∫ eˣ dx = eˣ

5. Substitute and Add the Constant of Integration:

   ∫ x eˣ dx = x eˣ - eˣ + C

   ∫ x eˣ dx = eˣ(x - 1) + C

Example 3: ∫ ln(x) dx

This example is interesting because it seems like there's only one function. However, we can consider it as ∫ ln(x) * 1 dx, where 1 is a constant function.

1. Choose u and dv:

   Following LIATE:

   • Logarithmic function: ln(x)
   • Algebraic function: 1

   Therefore, we choose:

   • u = ln(x)
   • dv = 1 dx

2. Find du and v:

   • du = (1/x) dx
   • v = ∫ 1 dx = x

3. Apply the Integration by Parts Formula:

   ∫ u dv = uv - ∫ v du

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

4. Evaluate the Remaining Integral:

   ∫ x (1/x) dx = ∫ 1 dx = x

5. Substitute and Add the Constant of Integration:

   ∫ ln(x) dx = x ln(x) - x + C

   ∫ ln(x) dx = x(ln(x) - 1) + C

Example 4: ∫ x² sin(x) dx

This example requires applying integration by parts twice.

1. First Application:

   • u = x²

   • dv = sin(x) dx

   • du = 2x dx

   • v = -cos(x)

   ∫ x² sin(x) dx = -x² cos(x) - ∫ (-cos(x)) (2x dx)

   ∫ x² sin(x) dx = -x² cos(x) + 2 ∫ x cos(x) dx

2. Second Application (for ∫ x cos(x) dx):

   • u = x

   • dv = cos(x) dx

   • du = dx

   • v = sin(x)

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

   ∫ x cos(x) dx = x sin(x) + cos(x) + C₁ (We don't need C₁ yet)

3. Substitute Back:

   ∫ x² sin(x) dx = -x² cos(x) + 2 [x sin(x) + cos(x)] + C

   ∫ x² sin(x) dx = -x² cos(x) + 2x sin(x) + 2 cos(x) + C

Example 5: ∫ eˣ cos(x) dx

This is a classic example where integration by parts is applied twice, and the original integral reappears.

1. First Application:

   • u = eˣ

   • dv = cos(x) dx

   • du = eˣ dx

   • v = sin(x)

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

2. Second Application (for ∫ sin(x) eˣ dx):

   • u = eˣ

   • dv = sin(x) dx

   • du = eˣ dx

   • v = -cos(x)

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

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

3. Substitute Back:

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

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

4. Solve for the Integral:

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

   ∫ eˣ cos(x) dx = (1/2) [eˣ sin(x) + eˣ cos(x)] + C

   ∫ eˣ cos(x) dx = (eˣ/2) [sin(x) + cos(x)] + C

Example 6: ∫ arctan(x) dx

Similar to the ln(x) example, we can rewrite this as ∫ arctan(x) * 1 dx.

1. Choose u and dv:

   Following LIATE:

   • Inverse trigonometric function: arctan(x)
   • Algebraic function: 1

   Therefore, we choose:

   • u = arctan(x)
   • dv = 1 dx

2. Find du and v:

   • du = (1/(1+x²)) dx
   • v = ∫ 1 dx = x

3. Apply the Integration by Parts Formula:

   ∫ u dv = uv - ∫ v du

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

4. Evaluate the Remaining Integral (using u-substitution):

   Let w = 1 + x² dw = 2x dx (1/2) dw = x dx

   ∫ x (1/(1+x²)) dx = (1/2) ∫ (1/w) dw = (1/2) ln|w| + C₁ = (1/2) ln(1+x²) + C₁

5. Substitute Back and Add the Constant of Integration:

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

Example 7: ∫ x³ ln(x) dx

1. Choose u and dv:

   Following LIATE:

   • Logarithmic function: ln(x)
   • Algebraic function: x³

   Therefore, we choose:

   • u = ln(x)
   • dv = x³ dx

2. Find du and v:

   • du = (1/x) dx
   • v = ∫ x³ dx = (x⁴/4)

3. Apply the Integration by Parts Formula:

   ∫ u dv = uv - ∫ v du

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

4. Evaluate the Remaining Integral:

   ∫ (x⁴/4) (1/x) dx = (1/4) ∫ x³ dx = (1/4) (x⁴/4) + C₁ = x⁴/16 + C₁

5. Substitute Back and Add the Constant of Integration:

   ∫ x³ ln(x) dx = (x⁴/4) ln(x) - x⁴/16 + C

   ∫ x³ ln(x) dx = (x⁴/16) (4 ln(x) - 1) + C

Example 8: ∫ arcsin(x) dx

Similar to arctan(x) and ln(x), we can rewrite this as ∫ arcsin(x) * 1 dx.

1. Choose u and dv:

   Following LIATE:

   • Inverse trigonometric function: arcsin(x)
   • Algebraic function: 1

   Therefore, we choose:

   • u = arcsin(x)
   • dv = 1 dx

2. Find du and v:

   • du = (1/√(1-x²)) dx
   • v = ∫ 1 dx = x

3. Apply the Integration by Parts Formula:

   ∫ u dv = uv - ∫ v du

   ∫ arcsin(x) dx = x arcsin(x) - ∫ x (1/√(1-x²)) dx

4. Evaluate the Remaining Integral (using u-substitution):

   Let w = 1 - x² dw = -2x dx (-1/2) dw = x dx

   ∫ x (1/√(1-x²)) dx = (-1/2) ∫ (1/√w) dw = (-1/2) ∫ w^(-1/2) dw = (-1/2) [2w^(1/2)] + C₁ = -√w + C₁ = -√(1-x²) + C₁

5. Substitute Back and Add the Constant of Integration:

   ∫ arcsin(x) dx = x arcsin(x) - (-√(1-x²)) + C

   ∫ arcsin(x) dx = x arcsin(x) + √(1-x²) + C

Example 9: Definite Integral: ∫₀^(π/2) x sin(x) dx

1. Apply Integration by Parts (as in Example 1):

   • u = x

   • dv = sin(x) dx

   • du = dx

   • v = -cos(x)

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

   ∫ x sin(x) dx = -x cos(x) + sin(x) + C (We don't need +C for definite integrals)

2. Evaluate the Definite Integral:

   ∫₀^(π/2) x sin(x) dx = [-x cos(x) + sin(x)] |₀^(π/2)

   = [-(π/2) cos(π/2) + sin(π/2)] - [-0 cos(0) + sin(0)]

   = [-(π/2) * 0 + 1] - [0 + 0]

   = 1

Example 10: ∫ x sec²(x) dx

1. Choose u and dv:

   Following LIATE:

   • Algebraic function: x
   • Trigonometric function: sec²(x)

   Therefore, we choose:

   • u = x
   • dv = sec²(x) dx

2. Find du and v:

   • du = dx
   • v = ∫ sec²(x) dx = tan(x)

3. Apply the Integration by Parts Formula:

   ∫ u dv = uv - ∫ v du

   ∫ x sec²(x) dx = x tan(x) - ∫ tan(x) dx

4. Evaluate the Remaining Integral:

   ∫ tan(x) dx = ∫ sin(x)/cos(x) dx

   Use u-substitution: Let w = cos(x), dw = -sin(x) dx

   ∫ tan(x) dx = -∫ (1/w) dw = -ln|w| + C₁ = -ln|cos(x)| + C₁

5. Substitute Back and Add the Constant of Integration:

   ∫ x sec²(x) dx = x tan(x) - (-ln|cos(x)|) + C

   ∫ x sec²(x) dx = x tan(x) + ln|cos(x)| + C

Tips and Tricks for Integration by Parts

• Choosing u and dv: The LIATE/ILATE rule is a good starting point, but consider the complexity of the resulting integral ∫ v du. Sometimes, a different choice might lead to a simpler integral.
• Repeated Integration by Parts: Some integrals require applying integration by parts multiple times until the integral simplifies. Keep track of your steps to avoid errors.
• Cyclic Integrals: In certain cases, like ∫ eˣ cos(x) dx, applying integration by parts twice leads back to the original integral. Solve for the integral algebraically.
• Definite Integrals: Remember to evaluate the uv term at the limits of integration when dealing with definite integrals.
• U-Substitution in Conjunction: Sometimes, after applying integration by parts, you might need to use u-substitution to solve the remaining integral.
• Practice: The key to mastering integration by parts is practice. Work through numerous examples to develop your intuition and problem-solving skills.

Conclusion

Integration by parts is a fundamental technique for evaluating integrals that involve the product of two functions. By carefully selecting u and dv and applying the integration by parts formula, you can transform complex integrals into simpler ones that can be solved more easily. The examples provided in this guide offer a comprehensive overview of the application of integration by parts in various scenarios. Remember to practice regularly and experiment with different choices of u and dv to develop your proficiency in this essential calculus skill. With consistent effort, you'll be able to confidently tackle a wide range of integration problems.