1 Integration Techniques

A structured guide to evaluating and approximating integrals using antiderivatives, the Fundamental Theorem of Calculus, substitution, algebraic and trigonometric techniques, and numerical rules.

Antiderivatives and basic formulas

Integration measures accumulated change and reverses differentiation when an exists. It can describe signed area, displacement, and other accumulated quantities.

An of ff is a function FF satisfying

F′(x)=f(x).F'(x)=f(x).

Because constants disappear under differentiation, all antiderivatives of ff have the form F(x)+CF(x)+C. This gives the :

∫f(x) dx=F(x)+C.\int f(x)\,dx=F(x)+C.

Important basic formulas include

∫xn dx=xn+1n+1+C(n≠−1),\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\qquad (n\ne -1),
∫1x dx=ln⁡∣x∣+C,∫ex dx=ex+C,\int \frac{1}{x}\,dx=\ln|x|+C,\qquad \int e^x\,dx=e^x+C,
∫cos⁡x dx=sin⁡x+C,∫sin⁡x dx=−cos⁡x+C.\int \cos x\,dx=\sin x+C,\qquad \int \sin x\,dx=-\cos x+C.

The constant-multiple and sum rules allow an integrand to be separated:

∫kf(x) dx=k∫f(x) dx,\int kf(x)\,dx=k\int f(x)\,dx,
∫[f(x)+g(x)] dx=∫f(x) dx+∫g(x) dx.\int [f(x)+g(x)]\,dx=\int f(x)\,dx+\int g(x)\,dx.

For example,

∫(3x2−4x+5) dx=x3−2x2+5x+C.\int (3x^2-4x+5)\,dx=x^3-2x^2+5x+C.

Differentiating the result is a direct check.

Takeaway: Find an , include CC for an , and differentiate the result to verify it.

Definite integrals and signed accumulation

A has limits of integration and produces a number:

∫abf(x) dx.\int_a^b f(x)\,dx.

It represents signed accumulation from x=ax=a to x=bx=b. Regions where the function is below the xx-axis contribute negatively, so signed accumulation can differ from total geometric area.

Useful properties are

∫aaf(x) dx=0,\int_a^a f(x)\,dx=0,
∫abf(x) dx=−∫baf(x) dx,\int_a^b f(x)\,dx=-\int_b^a f(x)\,dx,
∫acf(x) dx+∫cbf(x) dx=∫abf(x) dx,\int_a^c f(x)\,dx+\int_c^b f(x)\,dx=\int_a^b f(x)\,dx,

and

∫ab[kf(x)+g(x)] dx=k∫abf(x) dx+∫abg(x) dx.\int_a^b [kf(x)+g(x)]\,dx=k\int_a^b f(x)\,dx+\int_a^b g(x)\,dx.

A does not include CC, because any constant cancels when the endpoint values are subtracted.

Takeaway: Pay attention to orientation, signs, and whether the problem asks for signed accumulation or geometric area.

The Fundamental Theorem in action

The has two complementary parts.

First, an accumulation function has the original integrand as its derivative. If ff is continuous and

F(x)=∫axf(t) dt,F(x)=\int_a^x f(t)\,dt,

then

F′(x)=f(x).F'(x)=f(x).

The variable tt is a dummy variable that keeps the variable endpoint xx separate from the variable of integration. If the upper endpoint is itself a function, the chain rule is also required:

G(x)=∫au(x)f(t) dt⟹G′(x)=f(u(x))u′(x).G(x)=\int_a^{u(x)}f(t)\,dt \quad\Longrightarrow\quad G'(x)=f(u(x))u'(x).

For example,

ddx(∫1x21+t3 dt)=2x1+x6.\frac{d}{dx}\left(\int_1^{x^2}\sqrt{1+t^3}\,dt\right)=2x\sqrt{1+x^6}.

Second, a can be evaluated using any FF:

∫abf(x) dx=F(b)−F(a)=F(x)∣ab.\int_a^b f(x)\,dx=F(b)-F(a)=\left.F(x)\right|_a^b.

For instance,

∫02(x2+1) dx=[x33+x]02=143.\int_0^2(x^2+1)\,dx=\left[\frac{x^3}{3}+x\right]_0^2=\frac{14}{3}.

Takeaway: Differentiation undoes accumulation, while endpoint subtraction evaluates accumulated change.

Reversing the chain and product rules

is most effective when an integrand contains a composite expression together with its derivative. Set

u=g(x),du=g′(x) dx.u=g(x),\qquad du=g'(x)\,dx.

Then rewrite the entire integral in terms of uu, integrate, and substitute back if the integral is indefinite.

For example,

∫6x(3x2+4)4 dx\int 6x(3x^2+4)^4\,dx

uses u=3x2+4u=3x^2+4 and du=6x dxdu=6x\,dx. Therefore,

∫6x(3x2+4)4 dx=∫u4 du=u55+C=(3x2+4)55+C.\int 6x(3x^2+4)^4\,dx =\int u^4\,du =\frac{u^5}{5}+C =\frac{(3x^2+4)^5}{5}+C.

For a , either return to xx before applying the original limits or transform the limits:

∫abf(g(x))g′(x) dx=∫g(a)g(b)f(u) du.\int_a^b f(g(x))g'(x)\,dx =\int_{g(a)}^{g(b)}f(u)\,du.

is based on the product rule:

∫u dv=uv−∫v du.\int u\,dv=uv-\int v\,du.

Choose uu so that differentiating it simplifies the expression, and choose dvdv so that it can be integrated readily. A common guideline is LIATE: logarithmic, inverse trigonometric, algebraic, trigonometric, exponential.

For example, with u=xu=x and dv=cos⁡x dxdv=\cos x\,dx, we have du=dxdu=dx and v=sin⁡xv=\sin x. Thus,

∫xcos⁡x dx=xsin⁡x−∫sin⁡x dx=xsin⁡x+cos⁡x+C.\int x\cos x\,dx=x\sin x-\int\sin x\,dx=x\sin x+\cos x+C.

Takeaway: Use for a chain-rule pattern and for products or factors that simplify when differentiated.

Trigonometric integrals and identities

Trigonometric identities often turn a difficult trigonometric integral into a or a standard formula.

For

∫sin⁡mxcos⁡nx dx,\int \sin^m x\cos^n x\,dx,

save one factor of sin⁡x\sin x when mm is odd and convert the remaining even power using sin⁡2x=1−cos⁡2x\sin^2x=1-\cos^2x. When nn is odd, save one factor of cos⁡x\cos x and use cos⁡2x=1−sin⁡2x\cos^2x=1-\sin^2x. If both powers are even, use

sin⁡2x=1−cos⁡2x2,cos⁡2x=1+cos⁡2x2.\sin^2x=\frac{1-\cos 2x}{2}, \qquad \cos^2x=\frac{1+\cos 2x}{2}.

For example,

∫sin⁡3xcos⁡2x dx\int\sin^3x\cos^2x\,dx

can be rewritten as sin⁡x(1−cos⁡2x)cos⁡2x\sin x(1-\cos^2x)\cos^2x. With u=cos⁡xu=\cos x, the result is

−cos⁡3x3+cos⁡5x5+C.-\frac{\cos^3x}{3}+\frac{\cos^5x}{5}+C.

For tangent and secant, use

1+tan⁡2x=sec⁡2x.1+\tan^2x=\sec^2x.

An even power of sec⁡x\sec x suggests reserving sec⁡2x\sec^2x for u=tan⁡xu=\tan x, while an odd power of tan⁡x\tan x can suggest reserving sec⁡xtan⁡x\sec x\tan x for u=sec⁡xu=\sec x. Frequently used results include

∫tan⁡x dx=−ln⁡∣cos⁡x∣+C,\int\tan x\,dx=-\ln|\cos x|+C,
∫sec⁡x dx=ln⁡∣sec⁡x+tan⁡x∣+C.\int\sec x\,dx=\ln|\sec x+\tan x|+C.

Takeaway: Inspect parity first, preserve a factor that matches a derivative, and use identities to convert the remaining powers.

Radicals and rational functions

is designed for radicals with one of three quadratic patterns:

  • For a2−x2\sqrt{a^2-x^2}, use x=asin⁡θx=a\sin\theta and 1−sin⁡2θ=cos⁡2θ1-\sin^2\theta=\cos^2\theta.

  • For a2+x2\sqrt{a^2+x^2}, use x=atan⁡θx=a\tan\theta and 1+tan⁡2θ=sec⁡2θ1+\tan^2\theta=\sec^2\theta.

  • For x2−a2\sqrt{x^2-a^2}, use x=asec⁡θx=a\sec\theta and sec⁡2θ−1=tan⁡2θ\sec^2\theta-1=\tan^2\theta.

After integrating in θ\theta, use a right triangle or the original to return to xx.

For example, consider

∫dx9+x2.\int\frac{dx}{\sqrt{9+x^2}}.

Set x=3tan⁡θx=3\tan\theta. Then dx=3sec⁡2θ dθdx=3\sec^2\theta\,d\theta and 9+x2=3sec⁡θ\sqrt{9+x^2}=3\sec\theta, so

∫dx9+x2=∫sec⁡θ dθ=ln⁡∣sec⁡θ+tan⁡θ∣+C.\int\frac{dx}{\sqrt{9+x^2}} =\int\sec\theta\,d\theta =\ln|\sec\theta+\tan\theta|+C.

Since tan⁡θ=x/3\tan\theta=x/3 and sec⁡θ=x2+9/3\sec\theta=\sqrt{x^2+9}/3, the result is

∫dx9+x2=ln⁡∣x+x2+9∣+C.\int\frac{dx}{\sqrt{9+x^2}} =\ln\left|x+\sqrt{x^2+9}\right|+C.

applies to rational functions after polynomial division when the numerator degree is not less than the denominator degree. Decompose the rational function according to its factor types, solve for the unknown coefficients, and integrate each simpler term.

For example,

5x+1(x−1)(x+2)=2x−1+3x+2,\frac{5x+1}{(x-1)(x+2)}=\frac{2}{x-1}+\frac{3}{x+2},

so

∫5x+1(x−1)(x+2) dx=2ln⁡∣x−1∣+3ln⁡∣x+2∣+C.\int\frac{5x+1}{(x-1)(x+2)}\,dx =2\ln|x-1|+3\ln|x+2|+C.

Takeaway: Match each radical or denominator structure to its standard decomposition or pattern.

Numerical methods and technique selection

estimates a from function values. Divide [a,b][a,b] into nn subintervals using

Δx=b−an,xi=a+iΔx.\Delta x=\frac{b-a}{n}, \qquad x_i=a+i\Delta x.

The midpoint rule uses the center of each subinterval:

Mn=Δx∑i=1nf(xi−1+xi2).M_n=\Delta x\sum_{i=1}^n f\left(\frac{x_{i-1}+x_i}{2}\right).

The trapezoidal rule connects neighboring points with line segments:

Tn=Δx2[f(x0)+2f(x1)+2f(x2)+⋯+2f(xn−1)+f(xn)].T_n=\frac{\Delta x}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+\cdots+2f(x_{n-1})+f(x_n)\right].

Simpson’s rule uses parabolic approximations over pairs of subintervals and therefore requires nn to be even:

Sn=Δx3[f(x0)+4f(x1)+2f(x2)+4f(x3)+⋯+4f(xn−1)+f(xn)].S_n=\frac{\Delta x}{3}\left[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+\cdots+4f(x_{n-1})+f(x_n)\right].

For ∫01x2 dx\int_0^1x^2\,dx with the trapezoidal rule and n=4n=4, Δx=1/4\Delta x=1/4, and the estimate is

T4=1132=0.34375.T_4=\frac{11}{32}=0.34375.

The exact value is

∫01x2 dx=13≈0.33333,\int_0^1x^2\,dx=\frac{1}{3}\approx0.33333,

so the absolute error is approximately 0.010420.01042. Increasing nn generally improves the estimate, and Simpson’s rule is often more accurate for smooth functions.

When selecting a technique, simplify first, then look for a direct formula, , , trigonometric identities, , , or . Always check an by differentiating it and check definite-integral signs and endpoint values.

Takeaway: Use exact techniques when practical; use numerical rules when an exact is unavailable or unnecessary, and compare estimates with exact values when possible.