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.
If F(x)=∫f(x)dx, then
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
and let u=g(x), giving us
an integral we can solve more easily:
Then, we substitute g(x) back in to get our desired result of kf(g(x))+C.
When we make a substitution in a definite integral in the form
we need to remember that the bounds are from x=a to x=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.
Consider the integral
we can break the fraction into two parts:
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
but the integral of a derivative simply returns the original function, so we have
Replacing u′ and v′ with dxdu and dxdv respectively and rearranging terms gives us the result
which we use to perform integration by parts.
To perform integration by parts, we use
Another common way of writing this is
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.
We choose u=x and dv=ex, because ex is easy to integrate, and the derivative of x is just 1, which is simpler:
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.
We let u=x2 and v′=e2x because x2 gets simpler when differentiated, and we know how tom integrate e2x:
So
Now we have
So
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.
Let u=cos(2x) and v′=ex, then
So
This new integral J can be split by parts, so let u=−2sin2x and v′=ex, then
So
But this last integral is a multiple of the original I, specifically −4I:
From equation (1) we have
Substituting in (2) we find
So
Nice work completing Techniques of Integration, here's a quick recap of what we covered:
Exercises checked off
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.
If F(x)=∫f(x)dx, then
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
and let u=g(x), giving us
an integral we can solve more easily:
Then, we substitute g(x) back in to get our desired result of kf(g(x))+C.
When we make a substitution in a definite integral in the form
we need to remember that the bounds are from x=a to x=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.
Consider the integral
we can break the fraction into two parts:
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
but the integral of a derivative simply returns the original function, so we have
Replacing u′ and v′ with dxdu and dxdv respectively and rearranging terms gives us the result
which we use to perform integration by parts.
To perform integration by parts, we use
Another common way of writing this is
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.
We choose u=x and dv=ex, because ex is easy to integrate, and the derivative of x is just 1, which is simpler:
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.
We let u=x2 and v′=e2x because x2 gets simpler when differentiated, and we know how tom integrate e2x:
So
Now we have
So
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.
Let u=cos(2x) and v′=ex, then
So
This new integral J can be split by parts, so let u=−2sin2x and v′=ex, then
So
But this last integral is a multiple of the original I, specifically −4I:
From equation (1) we have
Substituting in (2) we find
So
Nice work completing Techniques of Integration, here's a quick recap of what we covered:
Exercises checked off