How to Solve Integrals Using Integration by Parts

Integration by parts is a technique in calculus used to evaluate integrals involving the product of two functions, based on the product rule for differentiation.

Derivation of the Integration by Parts Formula Using the Product Rule

Let $u(x)$ and $v(x)$ be functions of $x$, both differentiable on the domain under consideration.

By the product rule for differentiation,

$\dfrac{d}{dx} \left[ u(x)\, v(x) \right] = \dfrac{du}{dx}\, v(x) + u(x)\, \dfrac{dv}{dx}$

Integrate both sides with respect to $x$:

$\displaystyle \int \dfrac{d}{dx} \left[ u(x) v(x) \right] dx = \int \dfrac{du}{dx} v(x) dx + \int u(x) \dfrac{dv}{dx} dx$

The left side simplifies by the fundamental theorem of calculus:

$u(x) v(x) = \int \dfrac{du}{dx} v(x) dx + \int u(x) \dfrac{dv}{dx} dx$

Rearranging terms:

$\int u(x) \dfrac{dv}{dx} dx = u(x) v(x) - \int v(x) \dfrac{du}{dx} dx$

Assign $du = \dfrac{du}{dx} dx$, $dv = \dfrac{dv}{dx} dx$. The formula becomes:

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

This is the standard formula for integration by parts.

Application of the Integration by Parts Formula to Indefinite Integrals

To compute $\int u\, dv$, select $u$ and $dv$ such that $du$ and $v = \int dv$ are easily obtained. The choice of $u$ and $dv$ significantly affects the integral's simplification. On substitution: $u \rightarrow du$, $dv \rightarrow v$.

Careful selection is required so that $\int v\, du$ is more tractable than $\int u\, dv$.

ILATE Rule for Selection of $u$ in Integration by Parts

When deciding $u$ in $\int u\, dv$, the conventional ILATE order is followed for best tractability:

I: Inverse trigonometric functions

L: Logarithmic functions

A: Algebraic functions (polynomials, rational functions)

T: Trigonometric functions

E: Exponential functions

Select $u$ as the function highest in the ILATE precedence. $dv$ is then the remaining part of the integrand.

Integration by Parts Formula for Definite Integrals

For limits $a$ and $b$, integrating both sides within $[a, b]$:

$\displaystyle \int_a^b u\, dv = \left. u v \right|_a^b - \int_a^b v\, du$

Here, $\left. u v \right|_a^b = u(b)v(b) - u(a)v(a)$.

Worked Example: $\int x e^{2x}\, dx$

Given: $\displaystyle \int x e^{2x}\, dx$

Assign $u = x$, so $du = dx$.

Assign $dv = e^{2x}dx$, so $v = \int e^{2x}dx = \frac{1}{2}e^{2x}$.

Substitute into the formula:

$\int x e^{2x} dx = x \cdot \frac{1}{2}e^{2x} - \int \frac{1}{2}e^{2x}\, dx$

Evaluate the remaining integral:

$\int \frac{1}{2}e^{2x}\, dx = \frac{1}{2} \cdot \frac{1}{2}e^{2x} = \frac{1}{4}e^{2x}$

Combine the results:

$\int x e^{2x} dx = \frac{x}{2}e^{2x} - \frac{1}{4}e^{2x} + C$

This provides the antiderivative of $x e^{2x}$.

Worked Example: $\int \ln x \, dx$

Given: $\displaystyle \int \ln x \, dx$

Assign $u = \ln x$, $du = \dfrac{1}{x}dx$.

Assign $dv = dx$, $v = x$.

Apply the integration by parts formula:

$\int \ln x\, dx = x \ln x - \int x \cdot \dfrac{1}{x} dx$

Simplify the remaining integral:

$\int x \cdot \dfrac{1}{x} dx = \int 1\, dx = x$

Thus:

$\int \ln x\, dx = x \ln x - x + C$

This is the standard result for the integral of $\ln x$.

Integration by Parts with Multiple Applications: Example with $\int x^2 \sin x\, dx$

Given: $\displaystyle \int x^2 \sin x\, dx$

Assign $u = x^2$, $du = 2x\, dx$.

Assign $dv = \sin x\, dx$, $v = -\cos x$.

Apply the formula:

$\int x^2 \sin x\, dx = -x^2 \cos x + \int 2x \cos x\, dx$

Apply integration by parts again to $\int 2x \cos x\, dx$. This time, let $u_2 = 2x\, (du_2 = 2 dx)$, $dv_2 = \cos x\, dx (v_2 = \sin x)$:

$\int 2x \cos x\, dx = 2x \sin x - \int 2 \sin x\, dx$

$\int 2 \sin x\, dx = -2 \cos x$

Hence, combining results:

$\int 2x \cos x\, dx = 2x \sin x + 2 \cos x$

So, the complete answer is:

$\int x^2 \sin x\, dx = -x^2 \cos x + 2x \sin x + 2\cos x + C$

This demonstrates repeated application of integration by parts.

Tabular Method for Repeated Integration by Parts

For certain integrals, repeated application of integration by parts becomes algorithmic, especially with one algebraic and one trigonometric or exponential function. The tabular method involves writing derivatives of the algebraic function until zero and repeated antiderivatives of the other function; then, terms are combined with alternating signs.

A detailed explanation of tabular integration can be found within Integration By Parts Practice Paper.

Key Errors and Cues in Integration by Parts for Examinations

A common error is the incorrect or suboptimal choice of $u$ and $dv$, often making the resulting integral more complex than the original. Using the ILATE rule reduces this mistake.

When integrating by parts multiple times, failure to consistently apply signs or multiply through by constants carried from previous steps is a frequent source of computational error.

If, after two steps, the resulting integral indicates a recurrence of the original integrand, collect terms algebraically as shown in cyclic integrals to solve explicitly for the desired integral.

Further practice on products of standard functions can be found at Integral Calculus Important Questions.