CourseCourseQ BankQuestion BankDocsDocuments

P3.5 - Integration

Syllabus
2019
Topic
P3.5
Level
A2

Learning objectives

P3.5.111 To include integration of standard functions such as sin 3x, Integration of ekx,, sin kx, cos kx xn e5x, and their sums and differences. 2x.P3.5.2Integration by recognitionIntegration by recognition of For example, to include integration of tan x, sec2 2x. known derivatives to include Students are expected to be able to use trigonometric integrals of the form identities to integrate, for example, sin2 x, tan2 x, cos2 3x. ∫ f′(x) dx = ln (f(x)) + c and f(x) ∫ f′(x)[f(x)]n dx = [ f(x) ]n+1 + c n +1.

Reverse standard derivatives with the correct scale factor

Integrate a sum or difference term by term. For an exponential or trigonometric function with a linear inner expression, divide by the inner coefficient because differentiation would multiply by it.

Integrand Antiderivative
ekxe^{kx} ekx/ke^{kx}/k
xmx^m, m≠−1m\ne-1 xm+1/(m+1)x^{m+1}/(m+1)
1/x1/x ln⁡∣x∣\ln|x|
sin⁡(kx)\sin(kx) −cos⁡(kx)/k-\cos(kx)/k
cos⁡(kx)\cos(kx) sin⁡(kx)/k\sin(kx)/k
akxa^{kx} akx/(kln⁡a)a^{kx}/(k\ln a), a>0a>0, a≠1a\ne1

For example, ∫(4e−2x+3x−5sin⁡3x+2cos⁡4x)dx\int\left(4e^{-2x}+\frac3x-5\sin3x+2\cos4x\right)dx =−2e−2x+3ln⁡∣x∣+53cos⁡3x+12sin⁡4x+C.=-2e^{-2x}+3\ln|x|+\frac53\cos3x+\frac12\sin4x+C. Differentiating the result term by term restores the integrand.

For a definite integral, find one antiderivative FF and calculate F(b)−F(a)F(b)-F(a). The arbitrary constant cancels, so it is not included in the final definite value.

The power rule excludes m=−1m=-1; that case gives ln⁡∣x∣\ln|x|. Include +C+C for an indefinite integral, and divide by the complete inner coefficient, including its sign.

Recognise a derivative hidden inside an integrand

Integration by recognition reverses the chain rule. Identify a repeated inner function f(x), compare the remaining factor with its derivative, and correct only by a constant multiplier.

∫f′(x)f(x) dx=ln⁡∣f(x)∣+C,∫f′(x)[f(x)]n dx=[f(x)]n+1n+1+C(n≠−1)\int\frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+C,\qquad \int f'(x)[f(x)]^n\,dx=\frac{[f(x)]^{n+1}}{n+1}+C\quad(n\ne-1)

Since ddx(5x2+4)=10x\frac{d}{dx}(5x^2+4)=10x, ∫10x(5x2+4)1/2dx=23(5x2+4)3/2+C.\int10x(5x^2+4)^{1/2}dx=\frac23(5x^2+4)^{3/2}+C. Likewise, ∫6x3x2+5dx=ln⁡(3x2+5)+C.\int\frac{6x}{3x^2+5}dx=\ln(3x^2+5)+C.

Integrand Rewrite or recognise Integral
sec⁡2(2x)\sec^2(2x) derivative of tan⁡(2x)\tan(2x) is 2sec⁡2(2x)2\sec^2(2x) 12tan⁡(2x)+C\frac12\tan(2x)+C
tan⁡x\tan x sin⁡x/cos⁡x\sin x/\cos x −ln⁡∣cos⁡x∣+C-\ln|\cos x|+C
sin⁡2x\sin^2x (1−cos⁡2x)/2(1-\cos2x)/2 x/2−sin⁡2x/4+Cx/2-\sin2x/4+C
tan⁡2x\tan^2x sec⁡2x−1\sec^2x-1 tan⁡x−x+C\tan x-x+C
cos⁡2(3x)\cos^2(3x) (1+cos⁡6x)/2(1+\cos6x)/2 x/2+sin⁡6x/12+Cx/2+\sin6x/12+C

Known inverse derivatives also give ∫dxa2+x2=1aarctan⁡(xa)+C,qquad∫dxa2−x2=arcsin⁡(xa)+C,\int\frac{dx}{a^2+x^2}=\frac1a\arctan\left(\frac xa\right)+C,qquad \int\frac{dx}{\sqrt{a^2-x^2}}=\arcsin\left(\frac xa\right)+C, for a>0a>0 on their real domains.

The remaining multiplier must differ from f'(x) only by a constant; otherwise the pattern does not apply directly. Rewrite trigonometric squares before integrating, and differentiate the final expression to verify every factor and sign.