Evaluating Integrals- Techniques and Practice Problems

What Evaluating Integrals Actually Means

You have a function. You need to find its antiderivative. That's integration. Sounds simple. It isn't.

Integration is the inverse of differentiation. Where differentiation gives you the rate of change, integration gives you the accumulated value. The problem? There is no single method that works for everything. You need a toolkit of techniques, and you need to know which one to grab.

This guide covers the techniques that actually work, with practice problems you can sink your teeth into. No motivational quotes. Just math.

Basic Antiderivative Rules

Before you learn the fancy stuff, master these. They come up constantly.

• The power rule: ∫xⁿ dx = xⁿ⁺¹/(n+1) + C, except when n = -1
• ∫1/x dx = ln|x| + C
• ∫eˣ dx = eˣ + C
• ∫sin(x) dx = -cos(x) + C
• ∫cos(x) dx = sin(x) + C

The constant C is non-negotiable. You forgot it, you lost points. That's the rule.

U-Substitution: Your First Line of Attack

U-substitution reverses the chain rule. When you see a composite function, this is usually your move.

How It Works

You pick a "u" that simplifies the integrand. Then du replaces the differential. Solve for the integral in terms of u, integrate, then substitute back.

Example: ∫2x·cos(x²) dx

Let u = x². Then du = 2x dx.

The integral becomes ∫cos(u) du = sin(u) + C = sin(x²) + C.

Done. No guesswork if you picked the right u.

When U-Substitution Fails

If your chosen u and du don't appear together in the integrand, you made the wrong call. Try a different substitution or move to another technique.

Integration by Parts: For Products of Functions

When you have a product of two different types of functions (polynomial × trig, polynomial × exponential, etc.), integration by parts is often the answer.

The Formula

∫u dv = uv - ∫v du

Pick your u using LIATE (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential). Higher on the list gets the u. The rest goes with dv.

Example: ∫x·eˣ dx

u = x, dv = eˣ dx

du = dx, v = eˣ

∫x·eˣ dx = x·eˣ - ∫eˣ dx = x·eˣ - eˣ + C = eˣ(x - 1) + C

When You Need to Apply It Twice

Some integrals require two rounds. Others create a loop where the original integral reappears. When that happens, solve for the integral algebraically.

Trigonometric Integrals

Integrals with only trig functions follow patterns. The strategy depends on the powers involved.

Powers of Sine and Cosine

• If the power of sine is odd, factor out one sine and convert the rest to cosine using sin²x = 1 - cos²x
• If the power of cosine is odd, do the opposite
• If both powers are even, use half-angle identities

Example: ∫sin³(x)·cos²(x) dx

sin³(x) = sin²(x)·sin(x) = (1 - cos²x)·sin(x)

Let u = cos(x), du = -sin(x) dx

The integral becomes ∫(1 - u²)·u²·(-du) = -∫(u² - u⁴) du

= -(u³/3 - u⁵/5) + C = -cos³(x)/3 + cos⁵(x)/5 + C

Tangent and Secant

∫tan(x) dx = -ln|cos(x)| + C

∫sec(x) dx = ln|sec(x) + tan(x)| + C

For higher powers, use reduction formulas or rewrite using sec = 1/cos, tan = sin/cos.

Trigonometric Substitution: When Roots Appear

When you see √(a² - x²), √(a² + x²), or √(x² - a²), trig substitution often works.

• √(a² - x²) → x = a·sin(θ)
• √(a² + x²) → x = a·tan(θ)
• √(x² - a²) → x = a·sec(θ)

Example: ∫dx/√(4 - x²)

Let x = 2·sin(θ), dx = 2·cos(θ) dθ

√(4 - x²) = √(4 - 4sin²θ) = 2·cos(θ)

The integral becomes ∫(2·cos(θ) dθ)/(2·cos(θ)) = ∫dθ = θ + C

Back-substitute: θ = arcsin(x/2)

Answer: arcsin(x/2) + C

Partial Fractions: Rational Functions

When you have a rational function where the numerator's degree is lower than the denominator's, decompose into partial fractions.

Linear Factors

For each linear factor (ax + b), write A/(ax + b).

Repeated Linear Factors

For (ax + b)², write A/(ax + b) + B/(ax + b)².

Quadratic Factors

For irreducible quadratics, write (Ax + B)/(quadratic).

Example: ∫dx/(x² - 5x + 6)

x² - 5x + 6 = (x - 2)(x - 3)

1/[(x-2)(x-3)] = A/(x-2) + B/(x-3)

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

Setting x = 3: 1 = A(0) + B(1) → B = 1

Setting x = 2: 1 = A(-1) + B(0) → A = -1

The integral is ∫[-1/(x-2) + 1/(x-3)] dx = -ln|x-2| + ln|x-3| + C

= ln|(x-3)/(x-2)| + C

Comparing Integration Techniques

Practice Problems

Work through these before checking the solutions. That's the only way this sticks.

Problem 1

∫(3x² + 2x - 5) dx

Solution: Apply the power rule term by term.

= 3·x³/3 + 2·x²/2 - 5x + C = x³ + x² - 5x + C

Problem 2

∫x·e²ˣ dx

Solution: Integration by parts. u = x, dv = e²ˣ dx

du = dx, v = e²ˣ/2

= x·e²ˣ/2 - ∫e²ˣ/2 dx = x·e²ˣ/2 - e²ˣ/4 + C

= e²ˣ(2x - 1)/4 + C

Problem 3

∫dx/(x² + 4x + 5)

Solution: Complete the square first.

x² + 4x + 5 = (x+2)² + 1

Let u = x + 2, du = dx

∫du/(u² + 1) = arctan(u) + C = arctan(x + 2) + C

Problem 4

∫sin⁴(x) dx

Solution: Use power-reduction formulas.

sin⁴(x) = (sin²x)² = [(1 - cos(2x))/2]²

= (1 - 2cos(2x) + cos²(2x))/4

cos²(2x) = (1 + cos(4x))/2

sin⁴(x) = (1 - 2cos(2x) + (1 + cos(4x))/2)/4

= (3 - 4cos(2x) + cos(4x))/8

Integrate term by term:

= (3x)/8 - (4·sin(2x))/(2·8) + (sin(4x))/(4·8) + C

= 3x/8 - sin(2x)/4 + sin(4x)/32 + C

Common Mistakes That Cost You Points

• Dropping the constant of integration. Every indefinite integral needs +C.
• Forgetting to substitute back. Your answer must be in terms of the original variable.
• Wrong choice of technique. U-sub doesn't fix everything. Learn to recognize patterns.
• Sign errors in integration by parts. The formula is uv - ∫v du. The minus sign stays.
• Not completing the square. You can't partial fraction a non-factored quadratic.

Getting Started: A Practical Approach

Here's how to tackle any integral you encounter:

1. Look at the structure. Is it a product? A quotient? A root?
2. Check for obvious substitutions. Can you set u equal to something inside a composite function?
3. Identify function types. Polynomial + trig? Integration by parts. Rational function? Partial fractions. Root of quadratic? Trig substitution.
4. Try your approach. If it stalls after a few steps, try something else.
5. Verify by differentiating. Your antiderivative should give you back the original integrand.

Most integrals you'll see in a calculus course follow recognizable patterns. The more you practice, the faster pattern recognition gets. There's no shortcut here. You have to put in the work.