Integration techniques

Basic Integration Rules

Power Rule

\(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)\)

Exponential Functions

\(\int e^x \, dx = e^x + C\) \(\int a^x \, dx = \frac{a^x}{\ln a} + C\)

Logarithmic Functions

\(\int \ln x \, dx = x \ln x - x + C\) \(\int \log_a x \, dx = \frac{x \ln x - x}{\ln a} + C\)

Trigonometric Functions

\(\int \sin x \, dx = -\cos x + C\) \(\int \cos x \, dx = \sin x + C\) \(\int \tan x \, dx = -\ln|\cos x| + C = \ln|\sec x| + C\) \(\int \cot x \, dx = \ln|\sin x| + C\) \(\int \sec x \, dx = \ln|\sec x + \tan x| + C\) \(\int \csc x \, dx = \ln|\csc x - \cot x| + C\)

Inverse Trigonometric Functions

\(\int \frac{1}{\sqrt{1-x^2}} \, dx = \arcsin x + C\) \(\int \frac{1}{1+x^2} \, dx = \arctan x + C\) \(\int \frac{1}{|x|\sqrt{x^2-1}} \, dx = \text{arcsec } |x| + C\)

Hyperbolic Functions

\(\int \sinh x \, dx = \cosh x + C\) \(\int \cosh x \, dx = \sinh x + C\) \(\int \tanh x \, dx = \ln(\cosh x) + C\) \(\int \coth x \, dx = \ln|\sinh x| + C\)

Advanced Integration Techniques

1. U-Substitution (Substitution Method)

The substitution method works by replacing a complicated integral with a simpler one through a change of variable.

Procedure:

  1. Identify a substitution $u = g(x)$ such that $du = g’(x) \, dx$
  2. Rewrite the integral in terms of $u$
  3. Integrate with respect to $u$
  4. Substitute back to express the answer in terms of $x$

Example:

\(\int x \sin(x^2) \, dx\)

Let $u = x^2$, then $du = 2x \, dx$ or $x \, dx = \frac{du}{2}$

\[\int x \sin(x^2) \, dx = \int \sin(u) \frac{du}{2} = \frac{1}{2} \int \sin(u) \, du = -\frac{1}{2} \cos(u) + C = -\frac{1}{2} \cos(x^2) + C\]

2. Integration by Parts

Based on the product rule for derivatives, this technique is useful for integrating products of functions.

Formula:

\(\int u \, dv = uv - \int v \, du\)

where $u$ and $dv$ are chosen parts of the integrand.

Strategy for choosing $u$ and $dv$:

Remember the LIATE mnemonic (in descending priority):

  • L: Logarithmic functions
  • I: Inverse trigonometric functions
  • A: Algebraic functions
  • T: Trigonometric functions
  • E: Exponential functions

Choose $u$ from the earlier part of this list and $dv$ from the later part.

Example:

\(\int x \ln x \, dx\)

Let $u = \ln x$ and $dv = x \, dx$, then $du = \frac{1}{x} \, dx$ and $v = \frac{x^2}{2}$

\[\int x \ln x \, dx = \ln x \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \frac{x^2 \ln x}{2} - \int \frac{x}{2} \, dx = \frac{x^2 \ln x}{2} - \frac{x^2}{4} + C\]

3. Partial Fractions

This technique is used for integrating rational functions (ratios of polynomials).

Procedure:

  1. Ensure the numerator’s degree is less than the denominator’s; otherwise, perform polynomial long division
  2. Factor the denominator into linear and irreducible quadratic factors
  3. Decompose the rational function into a sum of simpler fractions
  4. Integrate each simple fraction separately

For Distinct Linear Factors:

If the denominator is $(x-a)(x-b)…(x-k)$, then: \(\frac{P(x)}{(x-a)(x-b)...(x-k)} = \frac{A}{x-a} + \frac{B}{x-b} + ... + \frac{K}{x-k}\)

For Repeated Linear Factors:

If the denominator has $(x-a)^m$, then: \(\frac{P(x)}{(x-a)^m} = \frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + ... + \frac{A_m}{(x-a)^m}\)

For Irreducible Quadratic Factors:

If the denominator has $(x^2+px+q)$ (irreducible), then: \(\frac{P(x)}{x^2+px+q} = \frac{Ax+B}{x^2+px+q}\)

Example:

\(\int \frac{dx}{x(x-2)}\)

Decompose into partial fractions: \(\frac{1}{x(x-2)} = \frac{A}{x} + \frac{B}{x-2}\)

Multiply both sides by $x(x-2)$: \(1 = A(x-2) + Bx\) \(1 = Ax - 2A + Bx\) \(1 = (A+B)x - 2A\)

Comparing coefficients: $A+B = 0$ and $-2A = 1$, so $A = -\frac{1}{2}$ and $B = \frac{1}{2}$

Therefore: \(\int \frac{dx}{x(x-2)} = \int \left( \frac{-\frac{1}{2}}{x} + \frac{\frac{1}{2}}{x-2} \right) \, dx = -\frac{1}{2} \ln|x| + \frac{1}{2} \ln|x-2| + C = \frac{1}{2} \ln\left|\frac{x-2}{x}\right| + C\)

4. Trigonometric Substitution

This technique is useful for integrals involving $\sqrt{a^2-x^2}$, $\sqrt{a^2+x^2}$, or $\sqrt{x^2-a^2}$.

For $\sqrt{a^2-x^2}$:

Substitute $x = a\sin\theta$ where $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$ Then $dx = a\cos\theta \, d\theta$ and $\sqrt{a^2-x^2} = a\cos\theta$

For $\sqrt{a^2+x^2}$:

Substitute $x = a\tan\theta$ where $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ Then $dx = a\sec^2\theta \, d\theta$ and $\sqrt{a^2+x^2} = a\sec\theta$

For $\sqrt{x^2-a^2}$:

Substitute $x = a\sec\theta$ where $0 \leq \theta < \frac{\pi}{2}$ or $\pi \leq \theta < \frac{3\pi}{2}$ Then $dx = a\sec\theta\tan\theta \, d\theta$ and $\sqrt{x^2-a^2} = a\tan\theta$

Example:

\(\int \frac{dx}{\sqrt{1-x^2}}\)

Let $x = \sin\theta$, then $dx = \cos\theta \, d\theta$ and $\sqrt{1-x^2} = \cos\theta$

\[\int \frac{dx}{\sqrt{1-x^2}} = \int \frac{\cos\theta \, d\theta}{\cos\theta} = \int d\theta = \theta + C = \arcsin x + C\]

5. Completing the Square

This technique helps simplify integrals containing quadratic expressions in the denominator or within a square root.

Procedure:

  1. Rearrange the quadratic into the form $(x+h)^2 + k$
  2. Make an appropriate substitution based on this form
  3. Integrate the resulting expression

Example:

\(\int \frac{dx}{x^2+4x+5}\)

Complete the square: $x^2+4x+5 = (x^2+4x+4) + 1 = (x+2)^2 + 1$

Let $u = x+2$, then $du = dx$ and the integral becomes: \(\int \frac{dx}{x^2+4x+5} = \int \frac{du}{u^2+1} = \arctan(u) + C = \arctan(x+2) + C\)

6. Using Trigonometric Identities

This approach is helpful for integrals involving products of trigonometric functions.

Useful Identities:

  • $\sin^2 x = \frac{1 - \cos(2x)}{2}$
  • $\cos^2 x = \frac{1 + \cos(2x)}{2}$
  • $\sin x \cos x = \frac{\sin(2x)}{2}$
  • $\sin(a) \sin(b) = \frac{\cos(a-b) - \cos(a+b)}{2}$
  • $\cos(a) \cos(b) = \frac{\cos(a-b) + \cos(a+b)}{2}$
  • $\sin(a) \cos(b) = \frac{\sin(a-b) + \sin(a+b)}{2}$

Example:

\(\int \sin^2 x \, dx\)

Using the identity $\sin^2 x = \frac{1-\cos(2x)}{2}$: \(\int \sin^2 x \, dx = \int \frac{1-\cos(2x)}{2} \, dx = \frac{1}{2} \int (1-\cos(2x)) \, dx = \frac{x}{2} - \frac{\sin(2x)}{4} + C\)

7. Integration of Special Functions

Rational Functions of Sine and Cosine

Substitute $t = \tan(x/2)$ to transform $\sin x$ and $\cos x$:

  • $\sin x = \frac{2t}{1+t^2}$
  • $\cos x = \frac{1-t^2}{1+t^2}$
  • $dx = \frac{2dt}{1+t^2}$

Integrals with $\sqrt{ax^2+bx+c}$

Complete the square and use an appropriate trigonometric or hyperbolic substitution.

Steps for Solving Complex Integrals

  1. Identify the appropriate technique based on the form of the integrand
  2. Simplify the integrand if possible (using algebraic manipulations, trigonometric identities, etc.)
  3. Apply the chosen integration technique methodically
  4. Verify your answer by differentiating it (to ensure it gives back the original integrand)

Worked Examples

Example 1: Using Substitution

Compute the antiderivative of $f(x) = \frac{\sqrt{x+1}}{2x+1}$

Solution: Let $u = x+1$, then $x = u-1$ and $dx = du$ The integrand becomes: \(\frac{\sqrt{u}}{2(u-1)+1} = \frac{\sqrt{u}}{2u-1}\)

This is still challenging, so let’s try a different approach. Let $v = \sqrt{x+1}$, then $v^2 = x+1$ and $x = v^2-1$ Also, $dx = 2v \, dv$

The integral becomes: \(\int \frac{\sqrt{x+1}}{2x+1} \, dx = \int \frac{v}{2(v^2-1)+1} \cdot 2v \, dv = \int \frac{2v^2}{2v^2-1} \, dv\) \(= \int \left(1 + \frac{1}{2v^2-1}\right) \, dv = v + \frac{1}{2} \int \frac{1}{v^2-\frac{1}{2}} \, dv\)

Continue with appropriate techniques to complete the integration.

Example 2: Integration by Parts

Compute the antiderivative of $f(x) = \frac{\sin x}{2\cos x + 1}$

Solution: This doesn’t immediately fit a standard integration by parts pattern. Let’s try substitution first.

Let $u = \cos x$, then $du = -\sin x \, dx$ or $\sin x \, dx = -du$ The integral becomes: \(\int \frac{\sin x}{2\cos x + 1} \, dx = -\int \frac{du}{2u + 1} = -\frac{1}{2} \int \frac{du}{u + \frac{1}{2}} = -\frac{1}{2} \ln\left|u + \frac{1}{2}\right| + C = -\frac{1}{2} \ln|2\cos x + 1| + C\)

Example 3: Partial Fractions

Compute the antiderivative of $f(x) = \frac{x+1}{x^2-4}$

Solution: Factor the denominator: $x^2-4 = (x-2)(x+2)$ Decompose into partial fractions: \(\frac{x+1}{(x-2)(x+2)} = \frac{A}{x-2} + \frac{B}{x+2}\)

Multiply both sides by $(x-2)(x+2)$: \(x+1 = A(x+2) + B(x-2)\) \(x+1 = Ax + 2A + Bx - 2B\) \(x+1 = (A+B)x + (2A-2B)\)

Comparing coefficients: $A+B = 1$ and $2A-2B = 1$, so $A = \frac{3}{4}$ and $B = \frac{1}{4}$

Therefore: \(\int \frac{x+1}{x^2-4} \, dx = \int \left( \frac{3/4}{x-2} + \frac{1/4}{x+2} \right) \, dx = \frac{3}{4} \ln|x-2| + \frac{1}{4} \ln|x+2| + C\)

Example 4: For functions with square roots

Compute $\int \frac{dx}{x^2+2}$

Solution: This can be solved by completing the square and using the arctangent integral formula.

\[\int \frac{dx}{x^2+2} = \int \frac{dx}{2((\frac{x}{\sqrt{2}})^2+1)} = \frac{1}{2}\int \frac{dx}{(\frac{x}{\sqrt{2}})^2+1}\]

Let $u = \frac{x}{\sqrt{2}}$, then $dx = \sqrt{2} \, du$ The integral becomes: \(\frac{1}{2}\int \frac{\sqrt{2} \, du}{u^2+1} = \frac{\sqrt{2}}{2}\int \frac{du}{u^2+1} = \frac{\sqrt{2}}{2} \arctan(u) + C = \frac{\sqrt{2}}{2} \arctan\left(\frac{x}{\sqrt{2}}\right) + C\)

See:

See

- Power Rule

- Antiderivatives

- Derivatives

- U-Substitution

- Integration by Parts

- Partial Fractions

Exercises:

  1. Compute the antiderivatives of $f(x) = \frac{\sqrt{x+1}}{2x+1}$
  2. Compute the antiderivatives of $f(x) = \frac{\sin x}{2\cos x + 1}$
  3. Compute the antiderivatives of $f(x) = \frac{x+1}{x^2-4}$
  4. Determine that $f(x) = \frac{x-2}{x^2+2}$ is not integrable using asymptotic comparison.