Integration Forms
Here follow the common integration formulas for a real-valued function. In each case,
understood to be a function of a single real variable.
1
. Integration: The General Power Formula We divide both sides by 4 so we can substitue into our original expression:
by M. Bourne
In this section, we apply this formula to trigonometric, logarithmic and
exponential functions:
Now to complete the required subsitution (u = sin-14x and the du/4 expression we
just found):
(We met this substitution formula in an earlier chapter: General Power Formula
for Integration.)
The expression on the right is a simple integral:
To complete the porblem, we substitute sin-14x for u:
Example 1: Integrate:
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Answer
We could either choose u = sin x, u = sin1/3x or u = cos x. However, only the first Example 3: Integrate:
one of these works in this problem.
So we let
u = sin x. Answer
Finding the differential: Let
du = cos x dx u = 3 + ln 2x
Substituting these into the integral gives: We can expand out the log term on the right hand side as follows:
3 + ln 2x = 3 + ln 2 + ln x
Now the first 2 terms on the right are constants (whose derivative equals zero)
and the derivative of the natural log of x is 1/x. So:
The last line is obtained by re-expressing our answer in terms of x.
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Example 2: Integrate:
Answer Example 4: Integrate:
Answer
Let
u = 1 − e-x
We have some choices for u in this example: sin-14x, 1 − 16x2, or √(1 − 16x2). The derivative of u is
Only one of these gives a result for du that we can use to integrate the given du/dx = 0 − (e-x) = e-x
expression, and that's the first one. So the differential du is:
So we let u = sin-14x. du = e-xdx
Then, using the derivative of the inverse sine, we have: We substitute to give:
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Example 5: Find the equation of the curve for which
if the curve passes through (1,2).
Answer
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2.
Answer
Therefore the equation is
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Exercises
Integrate each of the following functions:
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Answer
3.
Answer
Then
So we have the equation for y:
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6. A space vehicle is launched vertically from the ground such that its velocity v
(in km/s) is given by
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The graph of is as follows:
4.
Answer
Answer
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5. Find the equation of the curve for which
[We could use s or h in this problem.]
if the curve passes through (2,1). We need to find:
Answer Put u = ln(t3 + 1) and so
We need to find:
So
Now, since the height is 0 when t = 0, we have: K = 0.
The curve passes through (2, 1).
This means when x = 2, y = 1.
Introduction to derivative integral sPower formula:
formulas: (sin-1 x) =
In calculus, differentiation is an = nan-1
defined as the process of (cos-1 x) = -
finding the derivative which Product formula:
means measuring how a
function changes with respect (tan-1 x) =
to its input. (uv) = u' v + uv'
Integral calculus is a branch of Quotient formula:
(sec-1 x) =
calculus that deals with
integration. If ' f ' is a function
of real variable ' x ', then the =
definite integral is given by (cosec-1 x) = -
Reciprocal formula:
(cot-1 x) = -
=
Exponential / logarithmic derivative
Chain formula: formulas:
Where a and b are intervals of a real
line. f(g(x)) = f' (g(x)) g'(x) (ev) = ev
Following is the list of derivative and Inverse function formula:
integral formulas which help you for (av) = av(ln a)
learning derivative and integral
calculus.
f-1 (x) = (ln u) =
List of Derivative Formulas:
Trigonometric derivative formulas:
Constant function formula: (logx u) =
(sin x) = cos x Hyperbolic derivative formulas:
a = 0, where ' a ' is a constant
Scalar multiple formula: (cos x) = - sin x (sinh x) = cosh x
(au) = a , where ' a ' is a (tan x) = sec2 x (cosh x) = sinh x
constant
Sum formula: (cot x) = - cosec2x (tanh x) = sech2 x
(u + v) = + (sec x) = sec x tan x (coth x) = - cosech2x
Difference formula:
(cosec x) = - cosec x cot x (sech x) = - sech x tanh x
(u - v) = - Inverse trigonometric derivative
formulas: (cosech x) = - cosech x coth x
sList of Integral Formulas: ecx sin bx dx = (c sin bx - b tan x dx = ln |sec x| + C
cos bx) + C
Basic formulas of integral:
cot x dx = - ln |cosec x| + C
ecx cos bx dx = (c cos bx + b
sin bx) + C
xn dx = + C (for n ≠ - 1)
sec x dx = ln |sec x + tan x| + C
ln (x) dx = xln (x) - x + C
dx = ln |x| + C
Formulas of trigonometric integrals: cosec x dx = ln |cosec x - cot x| + C
u dv = uv - v du Formulas of hyperbolic integrals:
sin x dx = - cos x + C
Formulas of exponential / logarithmic
integrals: sinh x dx = cosh x + C
cos x dx = sin x + C
x x
e dx = e + C cosh x dx = sinh x + C
sec2 x dx = tan x + C
ecx dx = ecx + c tanh x dx = ln (cosh x) + C
cosec2 x dx = - cot x + C
acx dx = + C for a > 0, a ≠ 1 cosech x dx = ln |tanh | + C
sec x tan x dx = sec x + C
xecx dx = (cx - 1) + C sech x dx = arctan (sinh x) + C
cosec x cot x dx = - cosec x + C
coth x dx = ln |sinh x| + C
