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Techniques of Integration
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Techniques for evaluating integrals, including substitution for \(f(ax+b)\) and compositions, handling definite integrals with changed bounds, partial fractions, integration by parts, repeated parts, and cyclical parts.
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Integrating f(ax+b)
SL 5.10
If F(x)=∫f(x)dx, then
∫f(ax+b)dx=a1F(ax+b)🚫
Integration by substitution
SL 5.10
Integrating a composition of functions f(g(x)) requires us to divide by g′(x), so it is easier to find the anti-derivative of anything of the form g′(x)f′(g(x)) by first dividing by g′(x).
In symbols, we use the known fact
∫kg′(x)f′(g(x))dx=kf(g(x))+C🚫
and let u=g(x), giving us
∫kg′(x)f′(g(x))dx=k∫f′(u)du,🚫
an integral we can solve more easily:
k∫f′(u)du=kf(u)+C.
Then, we substitute g(x) back in to get our desired result of kf(g(x))+C.
Substitution and Integral Bounds
SL 5.10
When we make a substitution in a definite integral in the form
∫abkg′(x)f′(g(x))dx
we need to remember that the bounds are from x=a to x=b:
∫abkg′(x)f′(g(x))dx =k∫x=ax=bf′(u)du =[kf(u)]x=ax=b
We then have two choices:
Plug x=a and x=b into u to find the bounds in terms of u.
Plug u(x) back in and use the bounds a→b.
Integration by partial fractions
AHL 5.15
Consider the integral
∫(x−c)(x−d)ax+bdx,
we can break the fraction into two parts:
(x−c)(x−d)ax+b=x−cA+x−dB
Then, find A and B using a system of equations. Finally, integrate the two fractions.
Recall the product rule: (uv)′=u′v+uv′. Then, integrating both sides with respect to x yields
∫(uv)′dx=∫u′vdx+∫uv′dx,
but the integral of a derivative simply returns the original function, so we have
uv=∫u′vdx+∫uv′dx.
Replacing u′ and v′ with dxdu and dxdv respectively and rearranging terms gives us the result
∫udxdvdx=uv−∫vdxdudx,
which we use to perform integration by parts.
Integration by parts
AHL 5.16
To perform integration by parts, we use
∫udxdvdx=uv−∫vdxdudx📖
Another common way of writing this is
∫udv=uv−∫vdu📖
The first step is choosing what to make u and what to make dv. Since we will take the derivative of u, and integrate dv, we generally pick dv that is easy to integrate, and a u that gets simpler when we differentiate it.
Example: ∫xexdx
We choose u=x and dv=ex, because ex is easy to integrate, and the derivative of x is just 1, which is simpler:
u=x,dv=ex⇒uv=xex,vdu=ex
∫xexdx=xex−∫exdx=xex−ex+C
Repeated integration by parts
AHL 5.16
Certain functions, like powers of x, will simplify after repeated differentiation.
Therefore, we may perform repeated integration by parts using such functions as our u knowing that they will eventually yield a simple ∫vdu.
Example: I=∫x2e2xdx
We let u=x2 and v′=e2x because x2 gets simpler when differentiated, and we know how tom integrate e2x:
u′=2x,v=21e2x
So
I=x2⋅21e2x−J∫2x⋅21e2xdx
Now we have
J=∫xe2xdx,letu=x,v′=e2x
u′=1,v=21e2x
∴J=x⋅21e2x−∫1⋅21e2xdx=21e2x(x−21)
So
I=21e2x⋅(x2−x+21)+C
Cyclical integration by parts
AHL 5.16
Cyclical integration by parts are the result of a product of two functions that do not reduce via differentiation. Instead, they "cycle," or eventually return the original integral as the ∫vdu term.
In such cases, integrate by parts until you cycle back to the original integral I. Then, you may solve for I algebraically.
Example: I=∫cos(2x)exdx
Let u=cos(2x) and v′=ex, then
u′=−2sin2x,v=ex
So
I=cos(2x)ex−J∫(−2sin2x)exdx(1)
This new integral J can be split by parts, so let u=−2sin2x and v′=ex, then
u′=−4cos2x,v=ex
So
J=∫(−2sin2x)exdx=−2sin(2x)ex−∫(−4cos2x)exdx
But this last integral is a multiple of the original I, specifically −4I:
J=−2sin(2x)ex−(−4I)(2)
From equation (1) we have
I=cos(2x)ex−J
Substituting in (2) we find
I=cos(2x)ex+2sin(2x)ex+(−4I)
So
5I=ex(cos2x+2sin2x)⇒I=51ex(cos2x+2sin2x)+C
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