Explore basic tenets of integration (by parts, substitution method, and partial fractions)

Integration by parts reverses the product rule of differentiation. When the integrand is a product of two functions, this method separates them into u and dv, then applies:

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

Choosing u (ILATE Rule)

When selecting which part of the integrand becomes u, follow the priority order in the acronym ILATE:

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

The function higher in this list usually becomes u.

Worked Example 4

Find $\int x \sin x \, dx$.

Solution

Using ILATE, let $u = x$ (algebraic) and $dv = \sin x \, dx$ (trigonometric).

• $u = x$, so $du = dx$
• $dv = \sin x \, dx$, so $v = -\cos x$

Applying the formula: $\int x \sin x \, dx = uv - \int v \, du = x(-\cos x) - \int (-\cos x)(1) \, dx = -x\cos x + \int \cos x \, dx$

$= -x\cos x + \sin x + c$

Worked Example 5

Find $\int x^2 \ln x \, dx$.

Solution

Using ILATE, let $u = \ln x$ (logarithmic) and $dv = x^2 \, dx$ (algebraic).

• $u = \ln x$, so $du = \frac{1}{x} dx$
• $dv = x^2 \, dx$, so $v = \frac{x^3}{3}$

Applying the formula: $\int x^2 \ln x \, dx = \frac{x^3}{3}\ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} dx = \frac{x^3}{3}\ln x - \frac{1}{3}\int x^2 dx$

$= \frac{x^3}{3}\ln x - \frac{x^3}{9} + c$

Worked Example 6

Find $\int e^x \cos x \, dx$.

Solution

Here either function can be u. Let $u = e^x$ and $dv = \cos x \, dx$.

• $u = e^x$, so $du = e^x \, dx$
• $dv = \cos x \, dx$, so $v = \sin x$

First application: $\int e^x \cos x \, dx = e^x \sin x - \int e^x \sin x \, dx$

Now integrate $\int e^x \sin x \, dx$ similarly:

Let $u = e^x$ and $dv = \sin x \, dx$:

• $du = e^x \, dx$, $v = -\cos x$

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

Substitute back: $\int e^x \cos x \, dx = e^x \sin x - [-e^x \cos x + \int e^x \cos x \, dx]$

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

$2\int e^x \cos x \, dx = e^x(\sin x + \cos x)$

$\int e^x \cos x \, dx = \frac{1}{2}e^x(\sin x + \cos x) + c$

When the integrand is a rational function (a fraction with polynomials in numerator and denominator), we can decompose it into simpler fractions that are easier to integrate. This is particularly useful when the denominator can be factored.

Steps for Partial Fractions

1. Check if the fraction is proper (degree of numerator < degree of denominator). If not, perform polynomial long division first.
2. Factor the denominator completely
3. Set up partial fractions based on the factors
4. Solve for the unknown constants
5. Integrate each partial fraction separately

Worked Example 7

Find $\int \frac{2x}{x^2 - 4x + 3} \, dx$.

Solution

First, factor the denominator: $x^2 - 4x + 3 = (x - 1)(x - 3)$

Set up partial fractions: $\frac{2x}{(x-1)(x-3)} = \frac{A}{x-1} + \frac{B}{x-3}$

$2x = A(x-3) + B(x-1)$

To find A and B, substitute convenient values:

• Let $x = 1$: $2(1) = A(1-3) + B(0) \Rightarrow 2 = -2A \Rightarrow A = -1$
• Let $x = 3$: $2(3) = A(0) + B(3-1) \Rightarrow 6 = 2B \Rightarrow B = 3$

Thus: $\frac{2x}{(x-1)(x-3)} = \frac{-1}{x-1} + \frac{3}{x-3}$

Now integrate: $\int \frac{2x}{x^2 - 4x + 3} \, dx = \int \left(\frac{-1}{x-1} + \frac{3}{x-3}\right) dx = -\ln|x-1| + 3\ln|x-3| + c$

$= \ln\left|\frac{(x-3)^3}{x-1}\right| + c$

Worked Example 8

Find $\int \frac{5x - 2}{(x + 3)^2} \, dx$.

Solution

The denominator has a repeated linear factor. Set up: $\frac{5x - 2}{(x+3)^2} = \frac{A}{x+3} + \frac{B}{(x+3)^2}$

$5x - 2 = A(x+3) + B$

$5x - 2 = Ax + 3A + B$

Comparing coefficients:

• $A = 5$
• $3A + B = -2 \Rightarrow 15 + B = -2 \Rightarrow B = -17$

Thus: $\frac{5x - 2}{(x+3)^2} = \frac{5}{x+3} - \frac{17}{(x+3)^2}$

Integrate: $\int \frac{5x - 2}{(x+3)^2} \, dx = 5\int \frac{1}{x+3} dx - 17\int \frac{1}{(x+3)^2} dx = 5\ln|x+3| + \frac{17}{x+3} + c$

In Tanzania, these integration techniques are used in fields like engineering and construction. For example, when designing water storage tanks or grain silos, engineers calculate volumes of revolution to determine how much material is needed. A civil engineer might use integration by parts or substitution to find the exact volume of a cylindrical tank with domed ends—a shape common in rural water projects across regions like Mbeya and Arusha. Similarly, economists use these methods to calculate cumulative revenue or total cost from marginal functions, helping small businesses in local markets predict profits over time.