Integration of standard functions flashcards

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Integration by recognition
Spotting that the integrand is the derivative of a known function (often via the chain rule) and writing that function down, then adjusting any constant factor.
Definite integral ∫ₐᵇ f(x) dx
F(b) − F(a), where F is any integral of f. The constant of integration cancels.
Linear argument (inside function)
An inside function of the form ax + b, as in (ax + b)ⁿ, e^(ax + b) or sin(ax + b).
Exact value (of an integral)
An answer left in terms of ln, e, π or surds and simplified, rather than a calculator decimal.
Modulus in ln|x|
Makes the result valid for x < 0 as well as x > 0, because d/dx ln(−x) = 1/x too.
∫eˣ dx
eˣ + c (No — learn it)
∫1/x dx
ln|x| + c (No — learn it)
∫sin x dx and ∫cos x dx
−cos x + c and sin x + c (x in radians). (No — learn it)
∫sec²kx dx
(1/k) tan kx + c (Yes)
∫tan kx dx
(1/k) ln|sec kx| + c (Yes, for tan x; adjust for k)
∫f(ax + b) dx
(1/a) F(ax + b) + c, where F′ = f. Only for a linear inside function. (No — derive it)
∫(ax + b)ⁿ dx, n ≠ −1
(ax + b)ⁿ⁺¹ / (a(n + 1)) + c (No — derive it)
∫f′(x)/f(x) dx
ln|f(x)| + c (No — learn it)
∫f′(x)[f(x)]ⁿ dx, n ≠ −1
[f(x)]ⁿ⁺¹/(n + 1) + c (No — derive it)
sin²x and cos²x in terms of cos 2x
sin²x = ½(1 − cos 2x); cos²x = ½(1 + cos 2x) (No — learn it)
tan²x for integrating
tan²x = sec²x − 1, so ∫tan²x dx = tan x − x + c (No — learn it)
Area under y = 1/x from x = 1 to x = e
ln e − ln 1 = 1.
A region below the x-axis
Its definite integral is negative; the area is the size of that value.
Curve crosses the x-axis between the limits
Split the integral at each root and add the sizes of the parts.
Region between y = f(x) (upper) and y = g(x) (lower)
Area = ∫(f(x) − g(x)) dx between the x-coordinates of the intersections.
Graph of cos²x
Same as ½ + ½ cos 2x: a wave of period π about y = ½, between 0 and 1.
Why divide by a when integrating f(ax + b)?
Differentiating F(ax + b) multiplies by a (chain rule), so integrating must divide by a.
Why can't ∫(x² + 1)³ dx be done as (x² + 1)⁴/(4 × 2x)?
The divide-by-a rule only works for a linear inside; 2x is not a constant. Expand instead.
How do you test whether a fraction integrates to a logarithm?
Differentiate the denominator. If the numerator is a constant multiple of it, the answer is k ln|denominator|.
Why is ∫tan x dx = ln|sec x| + c?
tan x = sin x/cos x = −(−sin x)/cos x, which is −f′/f with f = cos x: −ln|cos x| = ln|sec x|.
Why can't you evaluate ∫₀¹ 1/(2x − 1) dx?
The integrand is undefined at x = ½, inside the interval (a vertical asymptote).
∫e^(2 − 3x) dx
−⅓e^(2 − 3x) + c.
∫5/(2x + 7) dx
(5/2) ln|2x + 7| + c.
∫sin 4x dx
−¼ cos 4x + c.
∫3x/(x² + 4) dx
3x = (3/2)(2x): (3/2) ln(x² + 4) + c.
∫cos x sin³x dx
f′f³ with f = sin x: ¼ sin⁴x + c.
∫₀^π sin²x dx
∫½(1 − cos 2x) dx = [x/2 − sin 2x/4] from 0 to π = π/2.
∫₁³ 2/x dx (exact)
[2 ln x] from 1 to 3 = 2 ln 3.
Writing ∫sin x dx = cos x + c
Wrong sign: d/dx(cos x) = −sin x, so ∫sin x dx = −cos x + c.
Writing ∫e^(3x) dx = 3e^(3x) + c
That is differentiation. Integrating divides by 3: ⅓e^(3x) + c.
Writing ∫cos²x dx = cos³x/3 + c
Use cos²x = ½(1 + cos 2x) first: x/2 + sin 2x/4 + c.
Using degrees in a trig integral
Calculus results for sin and cos hold only in radians.
Quoting a negative definite integral as an area
Give the size; if the curve crosses the axis, split at the roots first.

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