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Level 619

Integration by Parts


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∫u dv = uv - ∫ v du
If u and v are functions of x and have continuous derivatives, then
∫ u dv
= uv - ∫ v du
Then u will be the remaining factor(s) of the integrand.
Try letting dv be the most complicated portion of the integrand that fits a basic integration rule.
Then dv will be the remaining factor(s) of the integrand.
Try letting u be the portion of the integrand whose derivative is a function simpler than u.
Let u = x^n and let dv = e^ax dx, or cos ax dx.
For integrals of the form ∫x^n e^ax dx, ∫x^n sin ax dx, or ∫x^n arctan ax dx
Let u = ln x, arcsin ax, or arctan ax and let dv = x^n dx.
For integrals of the form ∫x^n ln x dx, ∫ x^n arcsin ax dx, or ∫ x^n arctan ax dx
Let u = sin bx or cos bx and let dv = e^ax dx.
For integrals of the form ∫ e^ax sin bx dx or ∫e^ax cos bx dx
the integration by parts formula
∫fg' dx=fg - ∫f'g dx
the integration by parts formula in words
the integral of f times g' is the same as f times g minus the integral of f' times g
u=
f(x)
du=
f'(x)dx
v=
g(x)
dv
g'(x)dx
v=∫dv
all we need to do is integrate dv
evaluate ∫xe^(6x) dx we choose
u=x, dv=e^(6x) dx, du=dx and v=∫e^(6x)dx = (1/6)e&(6x) dx
integration by parts definite integrals
∫from b to a u dv = uv|b to a - ∫ b to a v du
1
Sin^2(x) + cos^2(x)
Sin^2(x)
(1-cos(2x))/2
Cos^2(x)
(1+cos(2x))/2
If the power of cosine is odd and positive,
Save one cosine factor and convert the other factors to sine
If the power of sine is odd and positive,
Save one sine factor and convert other factors to cosine
Use: cos^2(x) = (1+cos(2x))/2
If the powers of both sine and cosine are even and positive,
Int(u) dv
uv - int(v) du
Sec^2(x)
1 + tan^2(x)
Csc^2(x)
-cotx+c
SinAcosB
1/2[sin(A-B) + sin(A+B)]
SinAsinB
1/2[sin(A-B) - cos(A+B)]
CosAcosB
1/2[cos(A-B) + cos(A+B)]
sinx/cosx
tanx= (quotient)
cosx/sinx
cotx= (quotient)
Whats the integration by parts formula?
∫u dv = uv - ∫v du
What does "convergent" mean?
If the limit exist, then we say that the improper integral converges.
What does "divergent" mean?
If the limit does not exist, then we say that the improper integral diverges.
An integral such that:
What is an improper integral?
What's L'Hospital's Rule and why is it useful?
L'Hospital's Rule is useful when an improper integral invivles a polynomial and an exponential. It states:
∫u dv=
uv-∫v du
∫xⁿe^(ax) dx
Take u=xⁿ & Tabular Method
∫xⁿ sin(ax) dx
Take u=xⁿ & Tabular Method
∫xⁿ cos(ax) dx
Take u=xⁿ & Tabular Method
∫xⁿ ln(x) dx
Take dv=xⁿ dx
∫xⁿ arcsin(ax) dx
Take dv=xⁿ dx
∫xⁿ arctan(ax) dx
Take dv=xⁿ dx
∫e^(ax) sin(bx) dx
Take dv=e^(ax) dx
∫e^(ax) cos(bx) dx
Take dv=e^(ax) dx