5.20 Integration by PartsHL ONLY

1. The formula for Integration by Parts

Integration by parts is the reverse process of the product rule for differentiation. Recall the product rule:

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

By integrating both sides with respect to $x$ and rearranging the terms, we elegantly obtain the formula for Integration by Parts:

$$\int u \dfrac{dv}{dx} dx = uv - \int v \dfrac{du}{dx} dx$$

Or, written in the more compact differential form:

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

2. Choosing $u$ and $dv$

The goal of integration by parts is to transform a complicated integral into a simpler one. To achieve this, we must choose our components $u$ and $dv$ wisely. A very helpful rule of thumb for choosing $u$ is the LIATE (or LIPET) acronym. You should systematically choose $u$ to be the function that appears first in the following list:

Logarithmic functions (e.g., $\ln x$, $\log_2 x$)
Inverse trigonometric functions (e.g., $\arctan x$, $\arcsin x$)
Algebraic functions (e.g., $x^2$, $3x$, $x^3-1$)
Trigonometric functions (e.g., $\sin x$, $\cos(2x)$)
Exponential functions (e.g., $e^x$, $e^{-3x}$)

The remaining portion of the integrand automatically becomes $dv$.

3. Basic Examples

Example 1: Algebraic and Exponential

Evaluate $\int x e^x dx$.

Following the LIATE rule, Algebraic ($x$) comes before Exponential ($e^x$).
Let $u = x \implies \dfrac{du}{dx} = 1 \implies du = dx$
Let $dv = e^x dx \implies v = \int e^x dx = e^x$
Applying the formula $\int u \, dv = uv - \int v \, du$ $$ \int x e^x dx = x e^x - \int e^x dx = \mathbf{x e^x - e^x + c} $$

Example 2: Logarithmic and Algebraic

Evaluate $\int x \ln x dx$.

Following the LIATE rule, Logarithmic ($\ln x$) comes before Algebraic ($x$).
Let $u = \ln x \implies du = \dfrac{1}{x} dx$
Let $dv = x dx \implies v = \int x dx = \dfrac{x^2}{2}$ $$ \begin{aligned} \int x \ln x dx &= (\ln x)\left(\dfrac{x^2}{2}\right) - \int \left(\dfrac{x^2}{2}\right) \left(\dfrac{1}{x}\right) dx \\ &= \dfrac{x^2}{2} \ln x - \int \dfrac{x}{2} dx \\ &= \mathbf{\dfrac{x^2}{2} \ln x - \dfrac{x^2}{4} + c} \end{aligned} $$

Example 3: Integrating a Single Function

Evaluate $\int \ln x dx$.

Trick: We can explicitly rewrite the integral as a product by multiplying by 1: $\int 1 \cdot \ln x dx$.
Let $u = \ln x \implies du = \dfrac{1}{x} dx$
Let $dv = 1 dx \implies v = x$ $$ \begin{aligned} \int \ln x dx &= x \ln x - \int x \left(\dfrac{1}{x}\right) dx \\ &= x \ln x - \int 1 dx \\ &= \mathbf{x \ln x - x + c} \end{aligned} $$

4. Repeated and Cyclic Integration by Parts

Example 4: Repeated Integration by Parts

Evaluate $\int x^2 \sin x dx$.

Let $u = x^2 \implies du = 2x dx$
Let $dv = \sin x dx \implies v = -\cos x$

$$ \begin{aligned} \int x^2 \sin x dx &= -x^2 \cos x - \int (-\cos x)(2x) dx \\ &= -x^2 \cos x + 2 \int x \cos x dx \end{aligned} $$ We apply integration by parts a second time for the integral $\int x \cos x dx$:
Let $u = x \implies du = dx$
Let $dv = \cos x dx \implies v = \sin x$ $$ \int x \cos x dx = x \sin x - \int \sin x dx = x \sin x + \cos x $$ Substituting this evaluation back into our main equation gives $$ \int x^2 \sin x dx = \mathbf{-x^2 \cos x + 2x \sin x + 2\cos x + c} $$

Example 5: The "Cyclic" Integral

Evaluate $I = \int e^x \cos x dx$.

Let $u = e^x \implies du = e^x dx$
Let $dv = \cos x dx \implies v = \sin x$
Then, $$ I = e^x \sin x - \int e^x \sin x dx $$ We apply integration by parts again for $\int e^x \sin x dx$:
Let $u = e^x \implies du = e^x dx$
Let $dv = \sin x dx \implies v = -\cos x$ $$ \begin{aligned} \int e^x \sin x dx &= -e^x \cos x - \int (-\cos x) e^x dx \\ &= -e^x \cos x + \int e^x \cos x dx \end{aligned} $$ Notice that the original integral $I$ has shown up again! We substitute this back into our equation $$ \begin{aligned} I &= e^x \sin x - \left(-e^x \cos x + I\right) \\ I &= e^x \sin x + e^x \cos x - I \\ 2I &= e^x (\sin x + \cos x) \\ I &= \mathbf{\dfrac{e^x}{2} (\sin x + \cos x) + c} \end{aligned} $$

5. Definite Integrals

Example 6: Definite Integral Evaluation

Evaluate the definite integral $\int_{0}^{\pi} x \cos x dx$.

From the previous techniques, we already established that the indefinite integral is $$ \int x \cos x dx = x \sin x + \cos x $$ We now evaluate this antiderivative across the precise boundaries from $0$ to $\pi$ $$ \begin{aligned} \int_{0}^{\pi} x \cos x dx &= \left[ x \sin x + \cos x \right]_{0}^{\pi} \\ &= (\pi \sin \pi + \cos \pi) - (0 \sin 0 + \cos 0) \\ &= (\pi(0) - 1) - (0 + 1) \\ &= -1 - 1 \\ &= \mathbf{-2} \end{aligned} $$

The exact graph of $f(x) = x \cos x$ visually confirms the accumulated net area. Notice that the negative region dictates the total sum since it is structurally larger than the positive lobe.