Run­ning dif­fer­en­ti­a­tion back­wards

Dif­fer­en­ti­a­tion takes a quan­tity and gives you its rate of change. Inte­gra­tion runs the film back­wards: you are handed the rate, and you have to recover the quan­tity. We will define the indef­i­nite inte­gral, col­lect the stan­dard inte­grals you should know by heart, and then learn two every­day meth­ods, sub­sti­tu­tion and the use of trigono­met­ric iden­ti­ties.

Anti­deriv­a­tives

FF is an anti­deriv­a­tive of ff if F′(x)=f(x)F'(x) = f(x). Any two anti­deriv­a­tives dif­fer only by a con­stant, which is why we always write

∫f(x) dx=F(x)+C.\displaystyle \int f(x)\,dx = F(x) + C.

Inte­gra­tion is also lin­ear, so you can split sums and pull out con­stants: ∫(af+bg) dx=a∫f dx+b∫g dx\displaystyle \int (af + bg)\,dx = a\int f\,dx + b\int g\,dx.

Stan­dard inte­grals

f(x)f(x) ∫f(x) dx\displaystyle \int f(x)\,dx
xnx^n (n≠−1n \ne -1) xn+1n+1\displaystyle \tfrac{x^{n+1}}{n + 1}
1x\displaystyle \tfrac{1}{x} log⁡∣x∣\log\lvert x \rvert
exe^x exe^x
axa^x axlog⁡a\displaystyle \tfrac{a^x}{\log a}
sin⁡x\sin x −cos⁡x-\cos x
cos⁡x\cos x sin⁡x\sin x
sec⁡2x\sec^2 x tan⁡x\tan x
csc⁡2x\csc^2 x −cot⁡x-\cot x
sec⁡xtan⁡x\sec x\tan x sec⁡x\sec x
csc⁡xcot⁡x\csc x\cot x −csc⁡x-\csc x
11−x2\displaystyle \tfrac{1}{\sqrt{1 - x^2}} sin⁡−1x\sin^{-1}x
11+x2\displaystyle \tfrac{1}{1 + x^2} tan⁡−1x\tan^{-1}x

Each of these car­ries its own +C+ C. Four more that come up con­stantly: ∫tan⁡x dx=log⁡∣sec⁡x∣\displaystyle \int\tan x\,dx = \log\lvert\sec x\rvert, ∫cot⁡x dx=log⁡∣sin⁡x∣\displaystyle \int\cot x\,dx = \log\lvert\sin x\rvert, ∫sec⁡x dx=log⁡∣sec⁡x+tan⁡x∣\displaystyle \int\sec x\,dx = \log\lvert\sec x + \tan x\rvert, ∫csc⁡x dx=log⁡∣csc⁡x−cot⁡x∣\displaystyle \int\csc x\,dx = \log\lvert\csc x - \cot x\rvert.

Exam­ple 1. ∫(4x3−3x+2ex)dx=x4−3log⁡∣x∣+2ex+C\displaystyle \int\left(4x^3 - \tfrac{3}{x} + 2e^x\right)dx = x^4 - 3\log\lvert x \rvert + 2e^x + C.

Exam­ple 2. ∫x2+3x−1x dx=∫(x3/2+3x1/2−x−1/2) dx=25x5/2+2x3/2−2x1/2+C\displaystyle \int\frac{x^2 + 3x - 1}{\sqrt{x}}\,dx = \int(x^{3/2} + 3x^{1/2} - x^{-1/2})\,dx = \tfrac{2}{5}x^{5/2} + 2x^{3/2} - 2x^{1/2} + C.

Exam­ple 3. Find ff with f′(x)=6x2−4f'(x) = 6x^2 - 4 and f(1)=3f(1) = 3. Inte­grat­ing, f(x)=2x3−4x+Cf(x) = 2x^3 - 4x + C, and the con­di­tion gives C=5C = 5.

Several curves y = 2x cubed - 4x + C, each a vertical shift of the others; the one with C = 5 is highlighted and passes through the point (1, 3).
All anti­deriv­a­tives are ver­ti­cal shifts of one another; the con­di­tion f(1) = 3 picks C = 5.

Sub­sti­tu­tion

Here is the sign to look for: the inte­grand con­tains some func­tion together with its deriv­a­tive. When you spot that, put tt = that func­tion, and the inte­gral usu­ally becomes one from the table.

Exam­ple 4. ∫2xcos⁡(x2) dx\displaystyle \int 2x\cos(x^2)\,dx. Put t=x2t = x^2, so dt=2x dxdt = 2x\,dx, and the answer is sin⁡(x2)+C\sin(x^2) + C.

Exam­ple 5. ∫etan⁡−1x1+x2 dx=etan⁡−1x+C\displaystyle \int\frac{e^{\tan^{-1}x}}{1 + x^2}\,dx = e^{\tan^{-1}x} + C.

Exam­ple 6. ∫(log⁡x)3x dx=(log⁡x)44+C\displaystyle \int\frac{(\log x)^3}{x}\,dx = \tfrac{(\log x)^4}{4} + C. ∫3x2x3+5 dx=log⁡∣x3+5∣+C\displaystyle \int\frac{3x^2}{x^3 + 5}\,dx = \log\lvert x^3 + 5 \rvert + C.

Exam­ple 7. ∫(3x+2)5 dx=(3x+2)618+C\displaystyle \int(3x + 2)^5\,dx = \tfrac{(3x + 2)^6}{18} + C.

Using trigono­met­ric iden­ti­ties

Exam­ple 8. Pow­ers of sine and cosine are best low­ered with dou­ble-angle for­mu­las. ∫sin⁡2x dx=∫1−cos⁡2x2 dx=x2−sin⁡2x4+C\displaystyle \int\sin^2 x\,dx = \int\tfrac{1 - \cos 2x}{2}\,dx = \tfrac{x}{2} - \tfrac{\sin 2x}{4} + C.

Exam­ple 9. A prod­uct of sines and cosines turns into a sum: ∫sin⁡5xcos⁡2x dx=12∫(sin⁡7x+sin⁡3x) dx=−cos⁡7x14−cos⁡3x6+C\displaystyle \int\sin 5x\cos 2x\,dx = \tfrac{1}{2}\int(\sin 7x + \sin 3x)\,dx = -\tfrac{\cos 7x}{14} - \tfrac{\cos 3x}{6} + C.

Exam­ple 10. ∫sin⁡3x dx=∫(1−cos⁡2x)sin⁡x dx=−cos⁡x+cos⁡3x3+C\displaystyle \int\sin^3 x\,dx = \int(1 - \cos^2 x)\sin x\,dx = -\cos x + \tfrac{\cos^3 x}{3} + C.

Try these your­self

  1. ∫(5x4+2x−7) dx\displaystyle \int(5x^4 + 2x - 7)\,dx; ∫(x+1x)2dx\displaystyle \int\left(\sqrt{x} + \tfrac{1}{\sqrt{x}}\right)^2dx; ∫(2sin⁡x−3sec⁡2x) dx\displaystyle \int(2\sin x - 3\sec^2 x)\,dx.
  2. Find ff if f′(x)=3cos⁡x−2exf'(x) = 3\cos x - 2e^x and f(0)=4f(0) = 4.
  3. ∫sin⁡(log⁡x)x dx\displaystyle \int\frac{\sin(\log x)}{x}\,dx; ∫cos⁡xsin⁡x dx\displaystyle \int\frac{\cos x}{\sqrt{\sin x}}\,dx; ∫xex2 dx\displaystyle \int xe^{x^2}\,dx.
  4. ∫2x+3x2+3x+7 dx\displaystyle \int\frac{2x + 3}{x^2 + 3x + 7}\,dx; ∫tan⁡3xsec⁡2x dx\displaystyle \int\tan^3 x\sec^2 x\,dx; ∫1xlog⁡x dx\displaystyle \int\frac{1}{x\log x}\,dx.
  5. ∫cos⁡23x dx\displaystyle \int\cos^2 3x\,dx; ∫sin⁡4xsin⁡2x dx\displaystyle \int\sin 4x\sin 2x\,dx; ∫cos⁡3x dx\displaystyle \int\cos^3 x\,dx.
  6. ∫dx1+cos⁡x\displaystyle \int\frac{dx}{1 + \cos x} (mul­ti­ply by 1−cos⁡x1 - \cos x).
  7. ∫sin⁡−1x1−x2 dx\displaystyle \int\frac{\sin^{-1}x}{\sqrt{1 - x^2}}\,dx.
  8. ∫sin⁡2xcos⁡2x dx\displaystyle \int\sqrt{\sin 2x}\cos 2x\,dx.

Answers to check against

Show answers
  1. x5+x2−7x+Cx^5 + x^2 - 7x + C; x22+2x+log⁡∣x∣+C\displaystyle \tfrac{x^2}{2} + 2x + \log\lvert x \rvert + C; −2cos⁡x−3tan⁡x+C-2\cos x - 3\tan x + C.
  2. 3sin⁡x−2ex+63\sin x - 2e^x + 6.
  3. −cos⁡(log⁡x)+C-\cos(\log x) + C; 2sin⁡x+C2\sqrt{\sin x} + C; ex22+C\displaystyle \tfrac{e^{x^2}}{2} + C.
  4. log⁡∣x2+3x+7∣+C\log\lvert x^2 + 3x + 7 \rvert + C; tan⁡4x4+C\displaystyle \tfrac{\tan^4 x}{4} + C; log⁡∣log⁡x∣+C\log\lvert\log x\rvert + C.
  5. x2+sin⁡6x12+C\displaystyle \tfrac{x}{2} + \tfrac{\sin 6x}{12} + C; sin⁡2x4−sin⁡6x12+C\displaystyle \tfrac{\sin 2x}{4} - \tfrac{\sin 6x}{12} + C; sin⁡x−sin⁡3x3+C\displaystyle \sin x - \tfrac{\sin^3 x}{3} + C.
  6. ∫1−cos⁡xsin⁡2xdx=−cot⁡x+csc⁡x+C\displaystyle \int\tfrac{1 - \cos x}{\sin^2 x}dx = -\cot x + \csc x + C.
  7. (sin⁡−1x)22+C\displaystyle \tfrac{(\sin^{-1}x)^2}{2} + C.
  8. (sin⁡2x)3/23+C\displaystyle \tfrac{(\sin 2x)^{3/2}}{3} + C.