Integration by Parts: Complete Guide + 10 Solved Problems

Integration by Parts is a powerful technique in calculus that allows us to integrate products of functions. It's essentially the inverse of the product rule for differentiation. When you encounter an integral that looks like a product of two different types of functions (e.g., a polynomial times an exponential, or a polynomial times a trigonometric function), integration by parts is often the key to solving it.

The Integration by Parts Formula

The formula for integration by parts is derived directly from the product rule:

If $u$ and $v$ are differentiable functions of $x$, then the product rule states:

$$\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$$

Integrating both sides with respect to $x$:

$$\int \frac{d}{dx}(uv) dx = \int u \frac{dv}{dx} dx + \int v \frac{du}{dx} dx$$

This simplifies to:

$$​uv = \int u \, dv + \int v \, du$$

Rearranging to solve for $\int u \, dv$, we get the Integration by Parts Formula:

$$\int u \, dv = uv - \int v \, du$$

The Strategy: Choosing $u$ and $dv$ (LIATE)

The success of integration by parts heavily depends on how you choose $u$ and $dv$. The goal is to make the new integral, $\int v \, du$, simpler to solve than the original integral, $\int u \, dv$. A helpful mnemonic for choosing $u$ is LIATE:

1. Logarithmic functions (e.g., $\ln x$)
2. Inverse trigonometric functions (e.g., $\arcsin x$, $\arctan x$)
3. Algebraic functions (e.g., $x^n$, polynomials)
4. Trigonometric functions (e.g., $\sin x$, $\cos x$)
5. Exponential functions (e.g., $e^x$, $a^x$)

Choose $u$ as the function that comes first in the LIATE order. The remaining part of the integrand will be $dv$. This rule generally works because functions earlier in LIATE simplify when differentiated, and functions later in LIATE are easy to integrate.

Step-by-Step Process

1. Choose $u$ and $dv$ from the integrand using the LIATE rule.
2. Differentiate $u$ to find $du$.
3. Integrate $dv$ to find $v$.
4. Apply the formula: $\int u \, dv = uv - \int v \, du$.
5. Evaluate the new integral ($\int v \, du$). If it's still complex, you might need to apply integration by parts again!

10 Solved Problems

Let's put the formula into practice with various examples.

Problem 1: Basic Example (Algebraic x Exponential)

Evaluate: $\int x e^x \, dx$

Solution:

1. Choose $u$ and $dv$: According to LIATE, Algebraic ($x$) comes before Exponential ($e^x$).
   • $u = x$
   • $dv = e^x \, dx$
2. Find $du$ and $v$:
   • $du = 1 \, dx$
   • $v = \int e^x \, dx = e^x$
3. Apply the formula: $\int u \, dv = uv - \int v \, du$ $$\int x e^x \, dx = x e^x - \int e^x (1) \, dx$$
4. Evaluate the new integral: $$\int x e^x \, dx = x e^x - e^x + C$$

Answer: $e^x(x - 1) + C$

Problem 2: Basic Example (Algebraic x Trigonometric)

Evaluate: $\int x \cos x \, dx$

Solution:

1. Choose $u$ and $dv$:
   • $u = x$
   • $dv = \cos x \, dx$
2. Find $du$ and $v$:
   • $du = 1 \, dx$
   • $v = \int \cos x \, dx = \sin x$
3. Apply the formula: $$\int x \cos x \, dx = x \sin x - \int \sin x \, dx$$
4. Evaluate the new integral: $$\int x \cos x \, dx = x \sin x - (-\cos x) + C$$ $$\int x \cos x \, dx = x \sin x + \cos x + C$$

Answer: $x \sin x + \cos x + C$

Problem 3: Involving Logarithmic Functions

Evaluate: $\int \ln x \, dx$

Solution:

1. Choose $u$ and $dv$: Logarithmic functions are first in LIATE. There's no other function, so we let $dv = dx$.
   • $u = \ln x$
   • $dv = dx$
2. Find $du$ and $v$:
   • $du = \frac{1}{x} \, dx$
   • $v = \int dx = x$
3. Apply the formula: $$\int \ln x \, dx = (\ln x)(x) - \int x \left(\frac{1}{x}\right) \, dx$$
4. Evaluate the new integral: $$\int \ln x \, dx = x \ln x - \int 1 \, dx$$ $$\int \ln x \, dx = x \ln x - x + C$$

Answer: $x \ln x - x + C$

Problem 4: Repeated Integration by Parts

Evaluate: $\int x^2 e^x \, dx$

Solution:

1. First application:
   • $u = x^2 \implies du = 2x \, dx$
   • $dv = e^x \, dx \implies v = e^x$
   $$\int x^2 e^x \, dx = x^2 e^x - \int e^x (2x) \, dx = x^2 e^x - 2 \int x e^x \, dx$$
2. Second application (on $\int x e^x \, dx$, which we solved in Problem 1):
   • We know $\int x e^x \, dx = x e^x - e^x + C$.
3. Substitute back: $$\int x^2 e^x \, dx = x^2 e^x - 2(x e^x - e^x) + C$$ $$\int x^2 e^x \, dx = x^2 e^x - 2x e^x + 2e^x + C$$

Answer: $e^x(x^2 - 2x + 2) + C$

Problem 5: Cycling Integrals (Requires solving for the integral)

Evaluate: $\int e^x \sin x \, dx$

Solution: This type of integral often requires two applications of integration by parts, leading back to the original integral, allowing you to solve for it algebraically.

1. First application:
   • $u = \sin x \implies du = \cos x \, dx$
   • $dv = e^x \, dx \implies v = e^x$
   $$I = \int e^x \sin x \, dx = e^x \sin x - \int e^x \cos x \, dx$$
2. Second application (on $\int e^x \cos x \, dx$):
   • $u = \cos x \implies du = -\sin x \, dx$
   • $dv = e^x \, dx \implies v = e^x$
   $$\int e^x \cos x \, dx = e^x \cos x - \int e^x (-\sin x) \, dx$$ $$\int e^x \cos x \, dx = e^x \cos x + \int e^x \sin x \, dx$$
3. Substitute back into the first equation: $$I = e^x \sin x - (e^x \cos x + \int e^x \sin x \, dx)$$ $$I = e^x \sin x - e^x \cos x - I$$
4. Solve for $I$: $$2I = e^x \sin x - e^x \cos x$$ $$I = \frac{1}{2} (e^x \sin x - e^x \cos x) + C$$

Answer: $\frac{1}{2} e^x (\sin x - \cos x) + C$

Problem 6: Definite Integral

Evaluate: $\int_0^{\pi/2} x \sin x \, dx$

Solution: First, find the indefinite integral $\int x \sin x \, dx$.

1. Choose $u$ and $dv$:
   • $u = x \implies du = dx$
   • $dv = \sin x \, dx \implies v = -\cos x$
2. Apply the formula: $$\int x \sin x \, dx = x(-\cos x) - \int (-\cos x) \, dx$$ $$= -x \cos x + \int \cos x \, dx$$ $$= -x \cos x + \sin x$$

Now, evaluate from $0$ to $\pi/2$:

$$[-x \cos x + \sin x]_0^{\pi/2} = \left(-\frac{\pi}{2} \cos \frac{\pi}{2} + \sin \frac{\pi}{2}\right) - (-0 \cos 0 + \sin 0)$$ $$= \left(-\frac{\pi}{2} \cdot 0 + 1\right) - (0 + 0)$$ $$= 1 - 0 = 1$$

Answer: $1$

Problem 7: Inverse Trigonometric Function

Evaluate: $\int \arctan x \, dx$

Solution:

1. Choose $u$ and $dv$:
   • $u = \arctan x \implies du = \frac{1}{1 + x^2} \, dx$
   • $dv = dx \implies v = x$
2. Apply the formula: $$\int \arctan x \, dx = x \arctan x - \int x \frac{1}{1 + x^2} \, dx$$
3. Evaluate the new integral (use substitution: let $w = 1 + x^2$, then $dw = 2x \, dx$): $$\int x \frac{1}{1 + x^2} \, dx = \frac{1}{2} \int \frac{1}{w} \, dw = \frac{1}{2} \ln|w| = \frac{1}{2} \ln(1 + x^2)$$
4. Substitute back: $$\int \arctan x \, dx = x \arctan x - \frac{1}{2} \ln(1 + x^2) + C$$

Answer: $x \arctan x - \frac{1}{2} \ln(1 + x^2) + C$

Problem 8: Algebraic and Logarithmic

Evaluate: $\int x^3 \ln x \, dx$

Solution:

1. Choose $u$ and $dv$:
   • $u = \ln x \implies du = \frac{1}{x} \, dx$
   • $dv = x^3 \, dx \implies v = \frac{x^4}{4}$
2. Apply the formula: $$\int x^3 \ln x \, dx = (\ln x)\left(\frac{x^4}{4}\right) - \int \frac{x^4}{4} \left(\frac{1}{x}\right) \, dx$$ $$= \frac{x^4}{4} \ln x - \int \frac{x^3}{4} \, dx$$
3. Evaluate the new integral: $$= \frac{x^4}{4} \ln x - \frac{1}{4} \frac{x^4}{4} + C$$ $$= \frac{x^4}{4} \ln x - \frac{x^4}{16} + C$$

Answer: $\frac{x^4}{4} \left(\ln x - \frac{1}{4}\right) + C$

Problem 9: Algebraic and Trigonometric (Repeated)

Evaluate: $\int x^2 \sin x \, dx$

Solution:

1. First application:
   • $u = x^2 \implies du = 2x \, dx$
   • $dv = \sin x \, dx \implies v = -\cos x$
   $$\int x^2 \sin x \, dx = -x^2 \cos x - \int (-\cos x)(2x) \, dx$$ $$= -x^2 \cos x + 2 \int x \cos x \, dx$$
2. Second application (on $\int x \cos x \, dx$, from Problem 2):
   • We know $\int x \cos x \, dx = x \sin x + \cos x$.
3. Substitute back: $$= -x^2 \cos x + 2(x \sin x + \cos x) + C$$ $$= -x^2 \cos x + 2x \sin x + 2 \cos x + C$$

Answer: $2x \sin x + (2 - x^2) \cos x + C$

Problem 10: Definite Integral with Cycling (Implicit Solving)

Evaluate: $\int_0^\pi e^{2x} \cos x \, dx$

Solution: First, find the indefinite integral $I = \int e^{2x} \cos x \, dx$.

1. First application:
   • $u = \cos x \implies du = -\sin x \, dx$
   • $dv = e^{2x} \, dx \implies v = \frac{1}{2} e^{2x}$
   $$I = \frac{1}{2} e^{2x} \cos x - \int \frac{1}{2} e^{2x} (-\sin x) \, dx$$ $$I = \frac{1}{2} e^{2x} \cos x + \frac{1}{2} \int e^{2x} \sin x \, dx$$
2. Second application (on $\int e^{2x} \sin x \, dx$):
   • $u = \sin x \implies du = \cos x \, dx$
   • $dv = e^{2x} \, dx \implies v = \frac{1}{2} e^{2x}$
   $$\int e^{2x} \sin x \, dx = \frac{1}{2} e^{2x} \sin x - \int \frac{1}{2} e^{2x} \cos x \, dx$$ $$= \frac{1}{2} e^{2x} \sin x - \frac{1}{2} I$$
3. Substitute back: $$I = \frac{1}{2} e^{2x} \cos x + \frac{1}{2} \left(\frac{1}{2} e^{2x} \sin x - \frac{1}{2} I\right)$$ $$I = \frac{1}{2} e^{2x} \cos x + \frac{1}{4} e^{2x} \sin x - \frac{1}{4} I$$
4. Solve for $I$: $$I + \frac{1}{4} I = \frac{1}{2} e^{2x} \cos x + \frac{1}{4} e^{2x} \sin x$$ $$\frac{5}{4} I = \frac{1}{4} e^{2x} (2 \cos x + \sin x)$$ $$I = \frac{1}{5} e^{2x} (2 \cos x + \sin x)$$

Now, evaluate the definite integral from $0$ to $\pi$:

$$\left[\frac{1}{5} e^{2x} (2 \cos x + \sin x)\right]_0^\pi$$ $$= \frac{1}{5} e^{2\pi} (2 \cos \pi + \sin \pi) - \frac{1}{5} e^{2(0)} (2 \cos 0 + \sin 0)$$ $$= \frac{1}{5} e^{2\pi} (2(-1) + 0) - \frac{1}{5} (1) (2(1) + 0)$$ $$= \frac{1}{5} e^{2\pi} (-2) - \frac{1}{5} (2)$$ $$= -\frac{2}{5} e^{2\pi} - \frac{2}{5}$$

Answer: $-\frac{2}{5} (e^{2\pi} + 1)$

Common Pitfalls and Tips

• Incorrect $u$ and $dv$ selection: Always use LIATE as a guideline. A bad choice can make the integral harder.
• Forgetting the $dx$: $du$ and $dv$ always include $dx$.
• Sign errors: Be careful with minus signs, especially when integrating trigonometric functions or when $v \, du$ has a negative.
• Recursion: For cycling integrals, remember to bring the integral to the left side and solve algebraically.
• Definite Integrals: Don't forget to evaluate the $uv$ term at the limits of integration.

With practice, integration by parts becomes a valuable tool in your calculus toolkit.