Integration By Parts Examples And Solutions

Integration by Parts: Practical Examples and Step‑by‑Step Solutions

Integration by parts is a staple technique in calculus, especially when dealing with products of functions where standard substitution fails. The formula, derived from the product rule for differentiation, states:

[ \int u , dv = uv - \int v , du ]

Choosing the right (u) and (dv) is crucial; the goal is to simplify the remaining integral ( \int v,du ). Below, we walk through a variety of examples—ranging from elementary to more advanced—to illustrate common strategies and pitfalls.

1. Classic Polynomial Times Exponential

Problem: (\displaystyle \int x e^{x} , dx)

Solution Steps

1. Choose (u) and (dv):

   • Let (u = x) (polynomial, easy to differentiate).
   • Let (dv = e^{x},dx) (exponential, easy to integrate).

2. Differentiate and Integrate:

   • (du = dx)
   • (v = e^{x})

3. Apply the Formula:
   [ \int x e^{x},dx = x e^{x} - \int e^{x},dx = x e^{x} - e^{x} + C ]

4. Factor if Desired:
   [ = e^{x}(x - 1) + C ]

Key Takeaway: When one factor is a polynomial, let it be (u); the integral usually simplifies.

2. Logarithm Times Power

Problem: (\displaystyle \int \ln(x),x^2 , dx)

Solution Steps

1. Set (u) and (dv):

   • (u = \ln(x)) (logarithm, derivative simplifies).
   • (dv = x^2,dx).

2. Compute (du) and (v):

   • (du = \frac{1}{x},dx)
   • (v = \frac{x^3}{3})

3. Apply the Formula:
   [ \int \ln(x),x^2 , dx = \frac{x^3}{3}\ln(x) - \int \frac{x^3}{3} \cdot \frac{1}{x},dx ] Simplify the remaining integral: [ = \frac{x^3}{3}\ln(x) - \frac{1}{3}\int x^2,dx ] [ = \frac{x^3}{3}\ln(x) - \frac{1}{3}\cdot \frac{x^3}{3} + C ] [ = \frac{x^3}{3}\ln(x) - \frac{x^3}{9} + C ]

Key Takeaway: Logarithms are excellent candidates for (u) because their derivatives simplify to rational functions.

3. Trigonometric Product

Problem: (\displaystyle \int x \sin(x) , dx)

Solution Steps

1. Choose (u = x), (dv = \sin(x),dx).

2. Differentiate/Integrate:

   • (du = dx)
   • (v = -\cos(x))

3. Apply the Formula:
   [ \int x \sin(x),dx = -x\cos(x) - \int -\cos(x),dx ] [ = -x\cos(x) + \int \cos(x),dx ] [ = -x\cos(x) + \sin(x) + C ]

Key Takeaway: For products of polynomials and trigonometric functions, let the polynomial be (u) Small thing, real impact..

4. Trigonometric Integral Requiring Repeated Parts

Problem: (\displaystyle \int \sin^2(x),dx)

Solution Steps

1. Rewrite Using Power‑Reducing Identity:
   [ \sin^2(x) = \frac{1 - \cos(2x)}{2} ] So the integral becomes: [ \int \frac{1}{2},dx - \int \frac{\cos(2x)}{2},dx ]

2. Integrate Directly:
   [ = \frac{x}{2} - \frac{1}{4}\sin(2x) + C ]

Alternate Approach (Integration by Parts):
Choose (u = \sin(x)), (dv = \sin(x),dx). After applying parts twice, you’ll arrive at the same result. Even so, the power‑reduction shortcut is simpler.

Key Takeaway: Some integrals are easier with trigonometric identities before applying parts Simple, but easy to overlook..

5. Exponential and Trigonometric Mix

Problem: (\displaystyle \int e^{x}\sin(x),dx)

Solution Steps

1. Set (u = e^{x}), (dv = \sin(x),dx).

2. Compute (du = e^{x},dx), (v = -\cos(x)).

3. First Application: [ \int e^{x}\sin(x),dx = -e^{x}\cos(x) - \int -e^{x}\cos(x),dx ] [ = -e^{x}\cos(x) + \int e^{x}\cos(x),dx ]

4. Second Application (for (\int e^{x}\cos(x),dx)):
   Let (u = e^{x}), (dv = \cos(x),dx).

   • (du = e^{x},dx), (v = \sin(x)).

   [ \int e^{x}\cos(x),dx = e^{x}\sin(x) - \int e^{x}\sin(x),dx ]

5. Substitute Back:
   Let (I = \int e^{x}\sin(x),dx).
   From step 3: [ I = -e^{x}\cos(x) + \int e^{x}\cos(x),dx ] From step 4: [ \int e^{x}\cos(x),dx = e^{x}\sin(x) - I ] Substitute into the expression for (I): [ I = -e^{x}\cos(x) + e^{x}\sin(x) - I ] Solve for (I): [ 2I = e^{x}(\sin(x) - \cos(x)) ] [ I = \frac{e^{x}}{2}(\sin(x) - \cos(x)) + C ]

Key Takeaway: When the integral recurs, set it equal to (I) and solve algebraically But it adds up..

6. Involving Logarithms and Trigonometric Functions

Problem: (\displaystyle \int \ln(x)\cos(x),dx)

Solution Steps

1. Choose (u = \ln(x)), (dv = \cos(x),dx).

2. Compute (du = \frac{1}{x},dx), (v = \sin(x)).

3. Apply the Formula: [ \int \ln(x)\cos(x),dx = \ln(x)\sin(x) - \int \sin(x)\cdot \frac{1}{x},dx ]

4. Remaining Integral:
   [ \int \frac{\sin(x)}{x},dx ] This is the sine integral function, often denoted as (\operatorname{Si}(x)). It does not have an elementary antiderivative.

5. Final Expression: [ = \ln(x)\sin(x) - \operatorname{Si}(x) + C ]

Key Takeaway: Not every integral yields an elementary function; recognize when a special function is required And it works..

7. Higher‑Degree Polynomial Times Logarithm

Problem: (\displaystyle \int x^3 \ln(x),dx)

Solution Steps

1. Set (u = \ln(x)), (dv = x^3,dx).

2. Compute (du = \frac{1}{x},dx), (v = \frac{x^4}{4}).

3. Apply the Formula: [ \int x^3 \ln(x),dx = \frac{x^4}{4}\ln(x) - \int \frac{x^4}{4}\cdot \frac{1}{x},dx ] Simplify: [ = \frac{x^4}{4}\ln(x) - \frac{1}{4}\int x^3,dx ] [ = \frac{x^4}{4}\ln(x) - \frac{1}{4}\cdot \frac{x^4}{4} + C ] [ = \frac{x^4}{4}\ln(x) - \frac{x^4}{16} + C ]

Key Takeaway: Even with higher powers, the pattern remains: let the logarithm be (u) Took long enough..

8. Integration Involving Inverse Trigonometric Functions

Problem: (\displaystyle \int \arcsin(x),dx)

Solution Steps

1. Set (u = \arcsin(x)), (dv = dx).

2. Compute (du = \frac{1}{\sqrt{1-x^2}},dx), (v = x).

3. Apply the Formula: [ \int \arcsin(x),dx = x\arcsin(x) - \int \frac{x}{\sqrt{1-x^2}},dx ]

4. Simplify the Remaining Integral:
   Let (w = 1 - x^2), (dw = -2x,dx).
   Then: [ \int \frac{x}{\sqrt{1-x^2}},dx = -\frac{1}{2}\int w^{-1/2},dw = -\frac{1}{2}\cdot 2w^{1/2} = -\sqrt{1-x^2} ]

5. Combine Results: [ = x\arcsin(x) + \sqrt{1-x^2} + C ]

Key Takeaway: Inverse trig functions work well as (u); the remaining integral often simplifies to a standard form Simple as that..

9. Repeated Integration by Parts: A Cubic Example

Problem: (\displaystyle \int x^3 e^{x},dx)

Solution Steps

1. First Application:

   • (u = x^3), (dv = e^{x},dx).
   • (du = 3x^2,dx), (v = e^{x}).
     [ I_1 = \int x^3 e^{x},dx = x^3 e^{x} - 3\int x^2 e^{x},dx ]

2. Second Application (for (\int x^2 e^{x},dx)):

   • (u = x^2), (dv = e^{x},dx).
   • (du = 2x,dx), (v = e^{x}).
     [ I_2 = \int x^2 e^{x},dx = x^2 e^{x} - 2\int x e^{x},dx ]

3. Third Application (for (\int x e^{x},dx)):

   • (u = x), (dv = e^{x},dx).
   • (du = dx), (v = e^{x}).
     [ I_3 = \int x e^{x},dx = x e^{x} - \int e^{x},dx = x e^{x} - e^{x} ]

4. Back‑Substitute: [ I_2 = x^2 e^{x} - 2(x e^{x} - e^{x}) = x^2 e^{x} - 2x e^{x} + 2e^{x} ] [ I_1 = x^3 e^{x} - 3\bigl(x^2 e^{x} - 2x e^{x} + 2e^{x}\bigr) ] [ = x^3 e^{x} - 3x^2 e^{x} + 6x e^{x} - 6e^{x} ] [ = e^{x}\bigl(x^3 - 3x^2 + 6x - 6\bigr) + C ]

Key Takeaway: Repeated parts can be organized systematically; each step reduces the polynomial degree by one Worth keeping that in mind..

10. FAQ: Common Pitfalls and Tips

This is the bit that actually matters in practice.

11. Conclusion

Integration by parts transforms a seemingly stubborn product into a simpler integral, provided the right pair of (u) and (dv) is chosen. Mastery comes from practice: recognize patterns, apply the LIATE rule, and don’t hesitate to iterate the process. Also, with the examples above, you now have a toolbox for tackling a wide range of integrals—from basic polynomials to mixed exponential‑trigonometric expressions. Keep experimenting, and soon the technique will become second nature No workaround needed..