16-NOV

Integration by parts <—> ‘%20%3D%20u’v%2Buv’#card=math&code=%28uv%29%27%20%3D%20u%27v%2Buv%27)

we integrate the derviative formula: ‘%20%3D%20%5Cint%20(u’v%20%2B%20uv’)#card=math&code=%5Cint%20%28uv%29%27%20%3D%20%5Cint%20%28u%27v%20%2B%20uv%27%29) LHS = ,

RHS = %20v(x)%20%2B%20u(x)v’(x))%20dx%20%20%3D%20%5Cint%20(u’(x)%20v(x))%20dx%20%2B%20%5Cint%20u(x)v’(x)%20dx%20%3D%20%5Cint%20v(x)du(x)%20%2B%20%5Cint%20u(x)dv(x)#card=math&code=%5Cint%20%28u%27%28x%29%20v%28x%29%20%2B%20u%28x%29v%27%28x%29%29%20dx%20%20%3D%20%5Cint%20%28u%27%28x%29%20v%28x%29%29%20dx%20%2B%20%5Cint%20u%28x%29v%27%28x%29%20dx%20%3D%20%5Cint%20v%28x%29du%28x%29%20%2B%20%5Cint%20u%28x%29dv%28x%29)

LHS = RHS —>

Formulation:

%20v’(x)dx%3D%20%20%5Cint%20u%20dv%20%3D%20uv%20-%20%5Cint%20v%20du%0A#card=math&code=%5Cint%20u%28x%29%20v%27%28x%29dx%3D%20%20%5Cint%20u%20dv%20%3D%20uv%20-%20%5Cint%20v%20du%0A)

Here, %2C%20v%3Dv(x)#card=math&code=u%3Du%28x%29%2C%20v%3Dv%28x%29) so %20dx%2C%20du%20%3D%20u’(x)%20dx#card=math&code=dv%20%3D%20v%27%28x%29%20dx%2C%20du%20%3D%20u%27%28x%29%20dx)

In full, the formula is %20v’(x)%20dx%20%3D%20u(x)v(x)%20-%20%5Cint%20v(x)%20u’(x)dx#card=math&code=%5Cint%20u%28x%29%20v%27%28x%29%20dx%20%3D%20u%28x%29v%28x%29%20-%20%5Cint%20v%28x%29%20u%27%28x%29dx)

How to apply this formula?

)%20g’(x)dx#card=math&code=%5Cint%20f%28g%28x%29%29%20g%27%28x%29dx) dx%20%3D%20d%20g(x)#card=math&code=g%27%28x%29dx%20%3D%20d%20g%28x%29)

%20%3D%20x%20(-%5Ccos%20x)%20-%20%5Cint%20(-%5Ccos%20)%20dx%20%3D%20-x%20%5Ccos%20x%20%2B%20%5Cint%20%5Ccos%20x%20dx%20%3D%20-x%20%5Ccos%20x%20%2B%20%5Csin%20x%20%2B%20C#card=math&code=%5Cint%20x%20%5Csin%20x%20dx%20%3D%20%5Cint%20x%20d%28-%5Ccos%20x%29%20%3D%20x%20%28-%5Ccos%20x%29%20-%20%5Cint%20%28-%5Ccos%20%29%20dx%20%3D%20-x%20%5Ccos%20x%20%2B%20%5Cint%20%5Ccos%20x%20dx%20%3D%20-x%20%5Ccos%20x%20%2B%20%5Csin%20x%20%2B%20C) now assign

$\int x^n \sin x d x= \int x^n d(-\cos x) = uv - \int v du = x^n (-\cos x) - \int (-\cos x) nx^{n-1} dx $ d

Now I assign %20%3D%20x%5E2%20e%5Ex%20-%20%5Cint%20e%5Ex%20(2x)dx#card=math&code=%5Cint%20x%5E2%20de%5Ex%20%3D%20uv%20-%20%5Cint%20v%20du%20%3D%20x%5E2%20e%5Ex%20-%20%5Cint%20e%5Ex%20d%28x%5E2%29%20%3D%20x%5E2%20e%5Ex%20-%20%5Cint%20e%5Ex%20%282x%29dx)

dx%20%3D%202%5Cint%20e%5Ex%20x%20dx#card=math&code=%5Cint%20e%5Ex%20%282x%29dx%20%3D%202%5Cint%20e%5Ex%20x%20dx) Now we assigne

%20%2BC%20%20%3D%20e%5Ex%20(%20x%5E2%20-%202x%20%2B%202)%2BC#card=math&code=%5Cint%20x%5E2%20e%5Ex%20dx%20%3D%20x%5E2%20e%5Ex%20-%202xe%5Ex%20%2B%202%28e%5Ex%29%20%2BC%20%20%3D%20e%5Ex%20%28%20x%5E2%20-%202x%20%2B%202%29%2BC)

Check: in formula 43, take .

(in formula 43, both .) =

(in formula 43, take ) $=e^x +C $

What if we do it another way? (that is, take to be the power function, instead of the exponential funciton)

$(1/3)\int e^x 3x^2 dx = (1/3) \int e^x d(x^3) =(1/3) e^x x^3 - (1/3) \int x^3 e^xdx $

Do it another way (but failed)

%5Cint%20%5Csin%20x%20(2xdx)%20%3D%20(1%2F2)%20%5Cint%20%5Csin%20x%20dx%5E2#card=math&code=%5Cint%20x%20%5Csin%20x%20dx%20%3D%20%281%2F2%29%5Cint%20%5Csin%20x%20%282xdx%29%20%3D%20%281%2F2%29%20%5Cint%20%5Csin%20x%20dx%5E2) and now let

$\int x \sin x dx = (1/2) x^2 \sin x - (1/2)\int x^2 \cos x dx $

%20x%5E2%20%5Csin%20x%20%20-%20(1%2F(2%5Ccdot%203))%20%20%5Cint%20%5Ccos%20x%20dx%5E3%20%3D%20(1%2F2)x%5E2%20%5Csin%20x%20-%201%2F6%20%5Bx%5E3%20%5Ccos%20x%20-%20%5Cint%20x%5E3%20d%20(%5Ccos%20x)%5D#card=math&code=%3D%20%281%2F2%29%20x%5E2%20%5Csin%20x%20%20-%20%281%2F%282%5Ccdot%203%29%29%20%20%5Cint%20%5Ccos%20x%20dx%5E3%20%3D%20%281%2F2%29x%5E2%20%5Csin%20x%20-%201%2F6%20%5Bx%5E3%20%5Ccos%20x%20-%20%5Cint%20x%5E3%20d%20%28%5Ccos%20x%29%5D)

priority (order) of appointing functions to be the : 1. trigonometric; 2. exponential; 3. powers; 4. logarithmic (inverse of exponential functions) ; 5. inverse trigonometric functions.

More substitutional techniques

1. involving of exponential functions.
   Ex. after Example 5.37. %20%5Cint%20(-6x%5E2)%20%5Cexp(-2x%5E3)%20dx%20%3D%20(-1%2F6)%20%5Cexp(-2x%5E3)%20%2BC#card=math&code=%5Cint%20x%5E2%20e%5E%7B-2x%5E3%7D%20dx%20%3D%20%28-1%2F6%29%20%5Cint%20%28-6x%5E2%29%20%5Cexp%28-2x%5E3%29%20dx%20%3D%20%28-1%2F6%29%20%5Cexp%28-2x%5E3%29%20%2BC)
   if we let %20%3D%20-2x%5E3#card=math&code=t%20%3Dt%28x%29%20%3D%20-2x%5E3), then )%20t’(x)%20dx%20%3D%20d%20%5B%5Cexp(t(x))%5D#card=math&code=%5Cexp%28t%28x%29%29%20t%27%28x%29%20dx%20%3D%20d%20%5B%5Cexp%28t%28x%29%29%5D)
   note that %20%3D%20-6%20x%5E2%20dx#card=math&code=t%27%28x%29%20%3D%20-6%20x%5E2%20dx)
   Let us try integration by parts.
   suppose we let #card=math&code=v%20%3D%20%5Cexp%28-2x%5E3%29) then %20(-6x%5E2)%20dx#card=math&code=dv%3D%20%5Cexp%28-2x%5E3%29%20%28-6x%5E2%29%20dx)
   we need the factor again: %20%5Cint%20(-6x%5E2)%20%5Cexp(-2x%5E3)%20dx%20%3D%20(-1%2F6)%20%5Cint%20u%20dv%20%3D%20(-1%2F6)%5Cint%20d%5B%5Cexp(-2x%5E3)%5D#card=math&code=%28-1%2F6%29%20%5Cint%20%28-6x%5E2%29%20%5Cexp%28-2x%5E3%29%20dx%20%3D%20%28-1%2F6%29%20%5Cint%20u%20dv%20%3D%20%28-1%2F6%29%5Cint%20d%5B%5Cexp%28-2x%5E3%29%5D) This means in the integration by parts formula, we have .
   Example 5.38 %5E%7B1%2F2%7D%20de%5Ex%20%3D%20%5Cint%20(e%5Ex%2B1)%5E%7B1%2F2%7D%20d(e%5Ex%2B1)%20%3D%20%5Cint%20u%5E%7B1%2F2%7D%20du%20%3D%20(3%2F2)%5E%7B-1%7D%20u%5E%7B3%2F2%7D%2BC#card=math&code=%5Cint%20e%5Ex%20%5Csqrt%7B1%2Be%5Ex%7D%20dx%20%3D%20%5Cint%20%28e%5Ex%2B1%29%5E%7B1%2F2%7D%20de%5Ex%20%3D%20%5Cint%20%28e%5Ex%2B1%29%5E%7B1%2F2%7D%20d%28e%5Ex%2B1%29%20%3D%20%5Cint%20u%5E%7B1%2F2%7D%20du%20%3D%20%283%2F2%29%5E%7B-1%7D%20u%5E%7B3%2F2%7D%2BC) also, %20%3D%20de%5Ex#card=math&code=d%281%2Be%5Ex%20%29%20%3D%20de%5Ex)
   The exp function has the nice property that ‘%20%3D%20e%5Ex#card=math&code=%28e%5Ex%29%27%20%3D%20e%5Ex) %20%3D%20e%5Ex%20dx#card=math&code=d%28e%5Ex%29%20%3D%20e%5Ex%20dx)

2. involving of trigonometric functions.

one is, we use trigonometric functions, to replace other functions, to simplify the integral.

the other one ,is the opposite: we use other functions (usually, rational functions) to replace trigonometric functions. (universal substitution for trigonometric functions)