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Integration Techniques






Beyond Direct Formulas


Most functions resist direct antidifferentiation. The integral of ex2e^{x^2} has no elementary formula. Even innocuous-looking expressions like 1+x3\sqrt{1 + x^3} lack closed-form antiderivatives.

Integration techniques transform difficult integrals into tractable ones. Substitution reverses the chain rule. Integration by parts reverses the product rule. Partial fractions decompose rational functions. Trigonometric methods handle roots and powers.

No single algorithm covers all cases—unlike differentiation, which follows systematic rules. Success requires recognizing which technique applies and executing it correctly. This pattern-matching skill develops through practice.



Why Techniques Are Needed


Differentiation follows mechanical rules: the derivative of any elementary function can be computed by applying chain, product, and quotient rules systematically.

Integration has no such universal algorithm. Some elementary functions have no elementary antiderivative. Others have antiderivatives that are difficult to find without insight.

Techniques bridge this gap. They transform integrands into forms matching known formulas. The transformation might involve changing variables, splitting products, or decomposing fractions—whatever reveals the underlying structure.

Substitution (u-Substitution)


Substitution reverses the chain rule. If the integrand contains a function and its derivative, substitution simplifies.

Method: Let u=g(x)u = g(x), so du=g(x)dxdu = g'(x)\, dx. Replace all xx-expressions with uu-expressions and integrate.

Example:

2xcos(x2)dx\int 2x \cos(x^2)\, dx


Let u=x2u = x^2, so du=2xdxdu = 2x\, dx:

=cosudu=sinu+C=sin(x2)+C= \int \cos u\, du = \sin u + C = \sin(x^2) + C


For definite integrals, convert the limits: when x=ax = a, u=g(a)u = g(a); when x=bx = b, u=g(b)u = g(b).

Integration by Parts


Integration by parts reverses the product rule:

udv=uvvdu\int u\, dv = uv - \int v\, du


Method: Identify factors uu and dvdv in the integrand. Differentiate uu to get dudu; integrate dvdv to get vv. Apply the formula.

Example:

xexdx\int x e^x\, dx


Let u=xu = x and dv=exdxdv = e^x\, dx. Then du=dxdu = dx and v=exv = e^x:

=xexexdx=xexex+C=ex(x1)+C= xe^x - \int e^x\, dx = xe^x - e^x + C = e^x(x - 1) + C


Choosing $u$: LIATE guides selection—Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. Earlier types typically make better choices for uu.

Trigonometric Integrals


Integrals involving powers of sine and cosine require strategic use of identities.

Odd power of sine: Save one sinx\sin x, convert remaining sin2x=1cos2x\sin^2 x = 1 - \cos^2 x, substitute u=cosxu = \cos x.

Odd power of cosine: Save one cosx\cos x, convert remaining cos2x=1sin2x\cos^2 x = 1 - \sin^2 x, substitute u=sinxu = \sin x.

Both powers even: Use half-angle identities:

sin2x=1cos2x2cos2x=1+cos2x2\sin^2 x = \frac{1 - \cos 2x}{2} \qquad \cos^2 x = \frac{1 + \cos 2x}{2}


Example:

sin3xdx=sinx(1cos2x)dx\int \sin^3 x\, dx = \int \sin x (1 - \cos^2 x)\, dx


Let u=cosxu = \cos x:

=(1u2)du=u+u33+C=cosx+cos3x3+C= -\int (1 - u^2)\, du = -u + \frac{u^3}{3} + C = -\cos x + \frac{\cos^3 x}{3} + C


Trigonometric Substitution


Square roots of quadratics suggest trigonometric substitutions.

For $\sqrt{a^2 - x^2}$: Let x=asinθx = a\sin\theta, so a2x2=acosθ\sqrt{a^2 - x^2} = a\cos\theta

For $\sqrt{a^2 + x^2}$: Let x=atanθx = a\tan\theta, so a2+x2=asecθ\sqrt{a^2 + x^2} = a\sec\theta

For $\sqrt{x^2 - a^2}$: Let x=asecθx = a\sec\theta, so x2a2=atanθ\sqrt{x^2 - a^2} = a\tan\theta

Example:

11x2dx\int \frac{1}{\sqrt{1 - x^2}}\, dx


Let x=sinθx = \sin\theta, so dx=cosθdθdx = \cos\theta\, d\theta and 1x2=cosθ\sqrt{1 - x^2} = \cos\theta:

=cosθcosθdθ=dθ=θ+C=arcsinx+C= \int \frac{\cos\theta}{\cos\theta}\, d\theta = \int d\theta = \theta + C = \arcsin x + C


Partial Fractions


Rational functions—polynomials divided by polynomials—decompose into simpler fractions.

Method: Factor the denominator. Write the fraction as a sum of terms with linear or irreducible quadratic denominators. Solve for coefficients. Integrate each term.

Example:

1x21dx=1(x1)(x+1)dx\int \frac{1}{x^2 - 1}\, dx = \int \frac{1}{(x-1)(x+1)}\, dx


Decompose:

1(x1)(x+1)=Ax1+Bx+1\frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}


Solving gives A=1/2A = 1/2, B=1/2B = -1/2:

=121x1dx121x+1dx=12lnx112lnx+1+C= \frac{1}{2}\int \frac{1}{x-1}\, dx - \frac{1}{2}\int \frac{1}{x+1}\, dx = \frac{1}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C


Choosing the Right Technique


Pattern recognition guides technique selection.

Substitution: Look for a function paired with its derivative. Expressions like f(g(x))g(x)f(g(x)) \cdot g'(x) signal substitution with u=g(x)u = g(x).

Parts: Products of different function types—polynomial times exponential, polynomial times trigonometric, logarithm times polynomial.

Trigonometric integrals: Powers of sinx\sin x and cosx\cos x, products of trigonometric functions.

Trigonometric substitution: Square roots of a2x2a^2 - x^2, a2+x2a^2 + x^2, or x2a2x^2 - a^2.

Partial fractions: Rational functions with factorable denominators.

Multiple techniques often combine. A problem might require substitution followed by parts, or partial fractions followed by a trigonometric integral.