~INTEGRATION FORMULAS & EXAMPLES~

1) ∫du = u + c

EXAMPLE: ∫dx = x + c

2) ∫k f(u)du = k∫f(u)du (k any constant)

EXAMPLE: ∫2pxdx = 2p∫xdx

3) ∫[f(u) + g(u)]du = ∫f(u)du + ∫g(u)du      (integrate a sum over each term)

EXAMPLE: ∫(x² + 2x + 3)dx = ∫x²dx + ∫2xdx + ∫3dx
                                               = ∫x²dx + 2∫xdx + 3∫dx

4) (POWER RULE): ∫uⁿdu = u⁽ⁿ⁺¹⁾ / (n+1) + C
(Must make sure du is present)

    EXAMPLE: ∫x²dx = x³ / 3 +C, since u=x, n=2, du=dx & is present.

    EXAMPLE: ∫(5x²+ 7)⁴xdx, since u=5 x²+7, du=10xdx (need to multiply inside by 10 & compensate by dividing on the outside by 10).

   (1/10)∫(5x²+7)⁴10xdx = (1/10)[(5x²+7)⁵]/5   + C = (5x²+7)⁵/50 + C

5) ∫e^udu = e^u+ C

EXAMPLE: ∫e^5xdx = (1/5)∫e^5x5dx = (1/5)e^5x+ C

6) ∫du/u = ln |u| + C

EXAMPLE: ∫xdx/(x² +7) = 1/2 ∫ 2xdx(x²+7) = (1/2)ln(x²+7)+C

EXAMPLE: ∫(sin3x)dx/cos3x = (-1/3)∫(-sin3x)3dx/cos3x
=(-1/3)ln |cos3x|+C

7) ∫cos(u)du = sin(u) +c

    EXAMPLE:   ∫ {[cos(lnx)] /x} dx = ∫ [cos(lnx)] [1/x] dx = sin(lnx) + C

8) ∫sin(u)du = -cos(u) +C

9) ∫sec²udu = tan(u) +C

10) ∫ sec(u)tan(u)du = sec(u) +C

EXAMPLE: ∫xsec(x²)tan(x²)dx = (1/2)∫ sec(x²)tan(x²) 2xdx
   = (1/2)sec(x²) + C

11) ∫ csc(u)cot(u)du = -csc(u) + C

12) ∫ csc²udu = -cot(u)+C

13) ∫ a^udu = a^u/ln(a) +C

14) ∫ du/(1+u²) = tan^-1(u) +C or -cot^-1u+C

EXAMPLE: ∫(xdx) /(1+x⁴) = (1/2)∫(2xdx)/ [1+(x²)²]
= (1/2) tan^-1(x²) + C

15) ∫ du /√(1-u²) = sin^-1u +C or -cos^-1u + C

16)  ∫ du / u√(u² - 1)= sec^-1|u|+c or -csc^-1|u|+C

Example of solving a differential equation by the "variables separable" method

Solve:     dy/dx = cos²y /sin²x →
dy /cos²y = dx /sin²x →
sec²y dy = csc²x dx →
∫secy² dy = ∫csc²x dx →
tan y = -cot x +C

    EXAMPLE: A particle moves in a straight line with acceleration a=12t²+6t. It starts from rest with an initial velocity of -3 ft/sec.
Find the equation for distance s at time t.

Since a=dv/dt, we have the differential equation   dv/dt = 12t²+6t. Separating the variables we get, dv = (12t²+6t)dt.
Integrating both sides, ∫ dv = ∫ (12t²+6t)dt. Which gives us,
v = 12t³/3 + 6t²/2 + C. Since v = -3 when t=0, C = -3.
Substitute -3 for C and we have, v = 4t³+3t²-3.
Since v = ds/dt, we have another diff. Eq., ds/dt=4t³+3t²-3.
Separating the variables & integrating, we get, s=t⁴+t³-3t+c.
Since the particle starts from rest, s=0 when t=0, so c=0.
Therefore, the equation for distance is, s = t⁴+t³-3t.

~NOTE: When separating the variables, make sure the differentials are "up stairs" and to the right of the expressions to be integrated.

~NOTE: An integral without an differential is meaningless.
In multivariable calculus, derivatives & integrals are taken with respect several different variables. Partial derivatives & multiple integrals are used. So, when differentiating, one must state reference to the variable. (on occasions, in single varaible calculus, it is understood).
However, not always. For example, the derivative of x wrt to x is 1, but    the derivative of x wrt to t is dx/dt. Likewise for integration. The differential to the far right of an integration gives the variable of integration. In multiple integrations (double & triple integrals), several different differentials are present. So, one would integrate wrt one, then integrate again, wrt another, & so on. These techniques are performed by treating all other varaibles as constants (temporally) until the operations are completed.