Techniques of Integration

Study Sheet

Techniques of Integration

Everything you need for AP Calculus BC, Topic 7

uu-Substitution Review

Concept
The idea: reverse the chain rule

If you can write an integrand in the form f(g(x)) g′(x)f(g(x))\,g'(x), let u=g(x)u=g(x) so that du=g′(x) dxdu=g'(x)\,dx. The integral becomes ∫f(u) du\displaystyle\int f(u)\,du. Look for an “inside” function whose derivative also appears (up to a constant factor). For a definite integral, either change the limits to uu-values or convert back to xx before substituting the original limits.

Reminder — The differentiation rules:  (xn)′=nxn−1,(fg)′=f′g+fg′,(fg)′=f′g−fg′g2,(f(g(x)))′=f′(g(x))g′(x)(x^{n})'=nx^{n-1},\quad (fg)'=f'g+fg',\quad \left(\tfrac{f}{g}\right)'=\frac{f'g-fg'}{g^{2}},\quad \big(f(g(x))\big)'=f'(g(x))g'(x)

Trickier substitutions to watch for: when only part of xx gets absorbed, you may need to solve for the leftover. For ∫xx−1 dx\displaystyle\int x\sqrt{x-1}\,dx, let u=x−1u=x-1, so x=u+1x=u+1 and dx=dudx=du.

Example
Solving for the leftover factor
∫xx−1 dx=∫(u+1)u du=∫(u3/2+u1/2) du=25u5/2+23u3/2+C=25(x−1)5/2+23(x−1)3/2+C.\begin{aligned} \int x\sqrt{x-1}\,dx &=\int (u+1)\sqrt{u}\,du=\int\big(u^{3/2}+u^{1/2}\big)\,du\\ &=\tfrac{2}{5}u^{5/2}+\tfrac{2}{3}u^{3/2}+C =\tfrac{2}{5}(x-1)^{5/2}+\tfrac{2}{3}(x-1)^{3/2}+C. \end{aligned}
Example
A substitution that clears a denominator
∫ln⁡xx dx,u=ln⁡x, du=1x dx  ⟹  ∫u du=12u2+C=12(ln⁡x)2+C.\int\frac{\ln x}{x}\,dx,\quad u=\ln x,\ du=\tfrac{1}{x}\,dx \;\Longrightarrow\;\int u\,du=\tfrac12 u^2+C=\tfrac12(\ln x)^2+C.
Tip

Tip: After substituting, every xx and dxdx must disappear. If an xx is left over, solve the uu-equation for xx and substitute it in --- don't guess.

Integration by Parts

Concept
The formula and LIATE

Integration by parts reverses the product rule:

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

Choose uu to be the factor that gets simpler when differentiated. The LIATE order tells you which factor to call uu (earlier = better choice for uu):

Logarithmic → Inverse trig → Algebraic → Trig → Exponential.\textbf{L}\text{ogarithmic}\ \to\ \textbf{I}\text{nverse trig}\ \to\ \textbf{A}\text{lgebraic}\ \to\ \textbf{T}\text{rig}\ \to\ \textbf{E}\text{xponential}.

Whatever is left (together with dxdx) becomes dvdv.

Reminder — Substitution and integration by parts:  ∫f(g(x))g′(x) dx=∫f(u) du,∫u dv=uv−∫v du\int f(g(x))g'(x)\,dx=\int f(u)\,du, \qquad \int u\,dv = uv-\int v\,du
Example
A full parts calculation

Evaluate ∫xex dx\displaystyle\int x e^{x}\,dx. By LIATE, Algebraic beats Exponential, so u=xu=x, dv=ex dxdv=e^{x}\,dx.

u=x⇒du=dx,dv=ex dx⇒v=ex.∫xex dx=xex−∫ex dx=xex−ex+C.\begin{aligned} u=x &\quad\Rightarrow\quad du=dx,\\ dv=e^{x}\,dx &\quad\Rightarrow\quad v=e^{x}.\\[2pt] \int x e^{x}\,dx &= xe^{x}-\int e^{x}\,dx = xe^{x}-e^{x}+C. \end{aligned}

Check: ddx(xex−ex)=ex+xex−ex=xex.\frac{d}{dx}\big(xe^x-e^x\big)=e^x+xe^x-e^x=xe^x. ✓

Repeated parts & the tabular method. When uu is a polynomial, you apply parts once per degree. The tabular method organizes this: differentiate uu down to 00 in one column, antidifferentiate dvdv in the other, and multiply along the diagonals with alternating signs +,−,+,…+,-,+,\dots

Example
Tabular method for ∫x2cos⁡x dx\displaystyle\int x^2\cos x\,dx
signD (u)I (dv)+x2cos⁡x−2xsin⁡x+2−cos⁡x−0−sin⁡x\begin{array}{c|cc} \text{sign} & D\ (u) & I\ (dv)\\\hline + & x^2 & \cos x\\ - & 2x & \sin x\\ + & 2 & -\cos x\\ - & 0 & -\sin x \end{array}

Multiply along diagonals with the signs:

∫x2cos⁡x dx=x2sin⁡x+2xcos⁡x−2sin⁡x+C.\int x^2\cos x\,dx = x^2\sin x + 2x\cos x - 2\sin x + C.

Cyclic parts. For ∫exsin⁡x dx\int e^x\sin x\,dx neither factor simplifies, but applying parts twice reproduces the original integral, which you solve for algebraically.

Example
Cyclic parts: ∫exsin⁡x dx\displaystyle\int e^{x}\sin x\,dx

Let I=∫exsin⁡x dxI=\int e^{x}\sin x\,dx. Take u=sin⁡x, dv=exdxu=\sin x,\ dv=e^x dx:  I=exsin⁡x−∫excos⁡x dx.\ I=e^x\sin x-\int e^x\cos x\,dx. Apply parts again to ∫excos⁡x dx\int e^x\cos x\,dx with u=cos⁡xu=\cos x:

I=exsin⁡x−(excos⁡x+∫exsin⁡x dx)I=exsin⁡x−excos⁡x−I2I=ex(sin⁡x−cos⁡x) ⇒ I=12ex(sin⁡x−cos⁡x)+C.\begin{aligned} I &= e^x\sin x-\Big(e^x\cos x+\int e^x\sin x\,dx\Big)\\ I &= e^x\sin x - e^x\cos x - I\\ 2I &= e^x(\sin x-\cos x)\ \Rightarrow\ I=\tfrac12 e^x(\sin x-\cos x)+C. \end{aligned}
Tip

Tip: For ∫ln⁡x dx\int \ln x\,dx and ∫arctan⁡x dx\int \arctan x\,dx there is no obvious dvdv --- take dv=dxdv=dx (so v=xv=x) and let uu be the whole function. Result: ∫ln⁡x dx=xln⁡x−x+C\int\ln x\,dx=x\ln x-x+C.

Rational Functions by Partial Fractions

Concept
Decomposing a proper rational function

To integrate P(x)Q(x)\dfrac{P(x)}{Q(x)} with deg⁡P<deg⁡Q\deg P<\deg Q: factor Q(x)Q(x), then write the fraction as a sum of simpler pieces.

  • [leftmargin=*,itemsep=1pt]
  • Each distinct linear factor (ax+b)(ax+b) contributes Aax+b\dfrac{A}{ax+b}.
  • A repeated linear factor (ax+b)k(ax+b)^k contributes A1ax+b+A2(ax+b)2+⋯+Ak(ax+b)k\dfrac{A_1}{ax+b}+\dfrac{A_2}{(ax+b)^2}+\cdots+\dfrac{A_k}{(ax+b)^k}.

If deg⁡P≥deg⁡Q\deg P\ge\deg Q, do polynomial long division first. Clear denominators, then solve for the constants (plug in convenient xx-values or match coefficients).

Example
Full partial-fraction integral (distinct linear factors)

Evaluate ∫5x−4x2−x−2 dx\displaystyle\int\frac{5x-4}{x^2-x-2}\,dx. Factor: x2−x−2=(x−2)(x+1)x^2-x-2=(x-2)(x+1).

5x−4(x−2)(x+1)=Ax−2+Bx+1  ⇒  5x−4=A(x+1)+B(x−2).\frac{5x-4}{(x-2)(x+1)}=\frac{A}{x-2}+\frac{B}{x+1} \;\Rightarrow\; 5x-4=A(x+1)+B(x-2).

Let x=2x=2: 6=3A⇒A=26=3A\Rightarrow A=2. Let x=−1x=-1: −9=−3B⇒B=3-9=-3B\Rightarrow B=3. So

∫ ⁣(2x−2+3x+1)dx=2ln⁡∣x−2∣+3ln⁡∣x+1∣+C.\int\!\Big(\frac{2}{x-2}+\frac{3}{x+1}\Big)dx = 2\ln|x-2|+3\ln|x+1|+C.
Example
A repeated linear factor
3x+1(x+1)2=Ax+1+B(x+1)2,3x+1=A(x+1)+B.\frac{3x+1}{(x+1)^2}=\frac{A}{x+1}+\frac{B}{(x+1)^2},\quad 3x+1=A(x+1)+B.

x=−1: B=−2.x=-1:\ B=-2. Matching xx: A=3.A=3. Then

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

Tip: Always check that the fraction is proper (deg⁡P<deg⁡Q\deg P<\deg Q) before decomposing. Otherwise divide first --- the quotient integrates as a polynomial and the remainder gets the partial fractions.

Trigonometric Integrals

Concept
Powers of sine and cosine

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

  • [leftmargin=*,itemsep=1pt]
  • If a power is odd, peel off one factor and convert the rest with sin⁡2x+cos⁡2x=1\sin^2x+\cos^2x=1, then substitute.
  • If both powers are even, use the power-reduction identities @@BLOCK0@@
Example
Odd power --- substitute
∫sin⁡3x dx=∫(1−cos⁡2x)sin⁡x dx=u=cos⁡x−∫(1−u2) du=−cos⁡x+13cos⁡3x+C.\int\sin^3 x\,dx=\int(1-\cos^2 x)\sin x\,dx \overset{u=\cos x}{=}-\int(1-u^2)\,du=-\cos x+\tfrac13\cos^3 x+C.
Example
Even power --- power reduction
∫sin⁡2x dx=∫1−cos⁡2x2 dx=x2−sin⁡2x4+C.\int\sin^2 x\,dx=\int\frac{1-\cos 2x}{2}\,dx=\frac{x}{2}-\frac{\sin 2x}{4}+C.
Tip

Tip: The double-angle identity sin⁡2x=2sin⁡xcos⁡x\sin 2x=2\sin x\cos x lets you rewrite ∫sin⁡xcos⁡x dx=12∫sin⁡2x dx=−14cos⁡2x+C\int\sin x\cos x\,dx=\tfrac12\int\sin 2x\,dx=-\tfrac14\cos 2x+C instantly.

Trigonometric Substitution (a first look)

Concept
Which substitution to use

When a root of a quadratic appears, trade xx for a trig function so the Pythagorean identity collapses the root:

a2−x2: x=asin⁡θ,a2+x2: x=atan⁡θ,x2−a2: x=asec⁡θ.\sqrt{a^2-x^2}:\ x=a\sin\theta,\qquad \sqrt{a^2+x^2}:\ x=a\tan\theta,\qquad \sqrt{x^2-a^2}:\ x=a\sec\theta.

After substituting, integrate in θ\theta, then use a right triangle to convert back to xx.

Reminder — The Pythagorean identity:  sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1
Example
∫dx9−x2\displaystyle\int\frac{dx}{\sqrt{9-x^2}} with x=3sin⁡θx=3\sin\theta

Then dx=3cos⁡θ dθdx=3\cos\theta\,d\theta and 9−x2=3cos⁡θ\sqrt{9-x^2}=3\cos\theta, so

∫3cos⁡θ3cos⁡θ dθ=∫dθ=θ+C=arcsin⁡ ⁣(x3)+C.\int\frac{3\cos\theta}{3\cos\theta}\,d\theta=\int d\theta=\theta+C=\arcsin\!\Big(\frac{x}{3}\Big)+C.
Tip

Tip: Many a2−x2\sqrt{a^2-x^2} and a2+x2\sqrt{a^2+x^2} integrals are faster with the standard arcsin/arctan formulas. Reach for a full trig sub when the root is multiplied by other powers of xx.

Improper Integrals

Concept
Rewrite as a limit

An integral is improper if a limit of integration is infinite, or if the integrand blows up somewhere on the interval. Handle it by replacing the bad endpoint with a limit:

∫a∞f(x) dx=lim⁡t→∞∫atf(x) dx,∫abf(x) dx=lim⁡t→b−∫atf(x) dx  (vertical asymptote at b).\int_a^{\infty}f(x)\,dx=\lim_{t\to\infty}\int_a^{t}f(x)\,dx,\qquad \int_a^{b}f(x)\,dx=\lim_{t\to b^-}\int_a^{t}f(x)\,dx\ \ (\text{vertical asymptote at }b).

If the limit is a finite number, the integral converges; if it is ±∞\pm\infty or does not exist, it diverges. Split an integral with two trouble spots into two pieces --- both must converge.

Example
Full improper integral (infinite limit) --- converges
∫1∞1x2 dx=lim⁡t→∞∫1tx−2 dx=lim⁡t→∞[−1x]1t=lim⁡t→∞(−1t+1)=0+1=1.(converges to 1)\begin{aligned} \int_1^{\infty}\frac{1}{x^2}\,dx &=\lim_{t\to\infty}\int_1^{t}x^{-2}\,dx =\lim_{t\to\infty}\Big[-\frac1x\Big]_1^{t}\\ &=\lim_{t\to\infty}\Big(-\frac1t+1\Big)=0+1=1.\quad(\textbf{converges to }1) \end{aligned}
Example
Unbounded integrand --- diverges

1x→∞\dfrac1x\to\infty as x→0+x\to0^+, so

∫011x dx=lim⁡t→0+[ln⁡x]t1=lim⁡t→0+(0−ln⁡t)=+∞.(diverges)\int_0^{1}\frac{1}{x}\,dx=\lim_{t\to0^+}\big[\ln x\big]_t^{1} =\lim_{t\to0^+}(0-\ln t)=+\infty.\quad(\textbf{diverges})
Tip

pp-integral test: ∫1∞dxxp\displaystyle\int_1^{\infty}\frac{dx}{x^{p}} converges   ⟺  p>1\iff p>1;   ∫01dxxp\displaystyle\int_0^{1}\frac{dx}{x^{p}} converges   ⟺  p<1\iff p<1.

Going Deeper: Advanced Techniques

Concept
1. Reduction formulas

Repeated integration by parts produces a formula that lowers a power by a fixed step. For example,

∫sin⁡nx dx=−sin⁡n−1xcos⁡xn+n−1n∫sin⁡n−2x dx.\int\sin^n x\,dx=-\frac{\sin^{n-1}x\cos x}{n}+\frac{n-1}{n}\int\sin^{n-2}x\,dx.

Apply it repeatedly until you reach ∫sin⁡x dx\int\sin x\,dx or ∫dx\int dx. The same pattern gives reduction formulas for cos⁡nx\cos^n x, sec⁡nx\sec^n x, xnexx^n e^x, and (ln⁡x)n(\ln x)^n.

Example
2. Deriving cyclic parts in general

For ∫eaxsin⁡(bx) dx\int e^{ax}\sin(bx)\,dx, two applications of parts give

∫eaxsin⁡(bx) dx=eax(asin⁡bx−bcos⁡bx)a2+b2+C.\int e^{ax}\sin(bx)\,dx=\frac{e^{ax}\big(a\sin bx-b\cos bx\big)}{a^2+b^2}+C.

The denominator a2+b2a^2+b^2 is the fingerprint of a cyclic “solve-for-II” problem.

Example
3. Partial fractions with an irreducible quadratic

An irreducible factor (x2+px+q)(x^2+px+q) contributes a linear numerator Bx+Cx2+px+q\dfrac{Bx+C}{x^2+px+q}. Evaluate:

2x+1x(x2+1)=Ax+Bx+Cx2+1.\frac{2x+1}{x(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}.

x=0: A=1.x=0:\ A=1. Matching: A+B=0⇒B=−1A+B=0\Rightarrow B=-1, and C=2C=2. So

∫2x+1x(x2+1) dx=ln⁡∣x∣−12ln⁡(x2+1)+2arctan⁡x+C,\int\frac{2x+1}{x(x^2+1)}\,dx=\ln|x|-\tfrac12\ln(x^2+1)+2\arctan x+C,

using ∫xx2+1dx=12ln⁡(x2+1)\int\frac{x}{x^2+1}dx=\tfrac12\ln(x^2+1) and ∫1x2+1dx=arctan⁡x\int\frac{1}{x^2+1}dx=\arctan x.

Concept
4. The comparison test for improper integrals

If 0≤f(x)≤g(x)0\le f(x)\le g(x) for all x≥ax\ge a, then:

∫a∞g dx converges ⇒ ∫a∞f dx converges;∫a∞f dx diverges ⇒ ∫a∞g dx diverges.\int_a^{\infty}g\,dx\ \text{converges}\ \Rightarrow\ \int_a^{\infty}f\,dx\ \text{converges};\qquad \int_a^{\infty}f\,dx\ \text{diverges}\ \Rightarrow\ \int_a^{\infty}g\,dx\ \text{diverges}.

Use it when an antiderivative is hard to find but the integrand is squeezed by a pp-integral.

Example
5. Comparison test in action

Does ∫1∞1x2+x dx\displaystyle\int_1^{\infty}\frac{1}{x^2+x}\,dx converge? For x≥1x\ge1,  x2+x>x2>0\ x^2+x> x^2>0, so 0<1x2+x<1x20<\frac{1}{x^2+x}<\frac{1}{x^2}. Since ∫1∞dxx2\int_1^{\infty}\frac{dx}{x^2} converges (a pp-integral with p=2>1p=2>1), the smaller integral converges by comparison.

Example
6. A full trig substitution with a leftover power

Evaluate ∫x24−x2 dx\displaystyle\int\frac{x^2}{\sqrt{4-x^2}}\,dx with x=2sin⁡θx=2\sin\theta, dx=2cos⁡θ dθdx=2\cos\theta\,d\theta, 4−x2=2cos⁡θ\sqrt{4-x^2}=2\cos\theta:

∫4sin⁡2θ2cos⁡θ 2cos⁡θ dθ=∫4sin⁡2θ dθ=2∫(1−cos⁡2θ) dθ=2θ−sin⁡2θ+C=2θ−2sin⁡θcos⁡θ+C=2arcsin⁡ ⁣(x2)−x4−x22+C,\begin{aligned} \int\frac{4\sin^2\theta}{2\cos\theta}\,2\cos\theta\,d\theta &=\int 4\sin^2\theta\,d\theta=2\int(1-\cos2\theta)\,d\theta\\ &=2\theta-\sin2\theta+C=2\theta-2\sin\theta\cos\theta+C\\ &=2\arcsin\!\Big(\frac{x}{2}\Big)-\frac{x\sqrt{4-x^2}}{2}+C, \end{aligned}

reading sin⁡θ=x2, cos⁡θ=4−x22\sin\theta=\frac{x}{2},\ \cos\theta=\frac{\sqrt{4-x^2}}{2} off the reference triangle.

Concept
7. Completing the square before substituting

When a quadratic has no rational roots, complete the square to expose a standard form. For ∫dxx2+2x+5\int\frac{dx}{x^2+2x+5}, write x2+2x+5=(x+1)2+4x^2+2x+5=(x+1)^2+4, then let u=x+1u=x+1:

∫duu2+4=12arctan⁡ ⁣(u2)+C=12arctan⁡ ⁣(x+12)+C.\int\frac{du}{u^2+4}=\frac12\arctan\!\Big(\frac{u}{2}\Big)+C=\frac12\arctan\!\Big(\frac{x+1}{2}\Big)+C.
Tip

Big picture: Before integrating, run a quick checklist --- (1) does a simple uu-sub work? (2) is it a product suited to parts? (3) is it a rational function (partial fractions)? (4) powers of trig? (5) a root of a quadratic (trig sub / complete the square)? Matching the form to the method is the whole game.

Formulas, Proofs & Tips

Tip
Substitution and integration by parts
∫f(g(x))g′(x) dx=∫f(u) du,∫u dv=uv−∫v du\int f(g(x))g'(x)\,dx=\int f(u)\,du, \qquad \int u\,dv = uv-\int v\,du

What it means. Substitution reverses the chain rule; parts reverses the product rule.

Example. ∫2x ex2 dx=ex2+C\int 2x\,e^{x^2}\,dx=e^{x^2}+C (let u=x2u=x^2, du=2x dxdu=2x\,dx).

Why it works. With u=g(x)u=g(x), du=g′(x)dxdu=g'(x)dx, so the integral literally rewrites in uu. For parts, integrate the product rule (uv)′=u′v+uv′(uv)'=u'v+uv' across the interval and rearrange.

Tip. When substituting in a definite integral, convert the limits to uu as well — then you never convert back.