Integration by parts

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\frac{d}{dx} \sin x=?\, This article/section deals with mathematical concepts appropriate for a student in late high school or early university.

This article details the method known as Integration by Parts.


Integration by Parts

Integration by parts is a special technique to facilitate the integration of the product of two functions that otherwise lack an obvious integral. This technique can be proven with the product rule.

The rule for integration by parts is stated as follows:[1]

\int f(x) g'(x)\,dx = f(x) g(x) - \int f'(x) g(x)\,dx,

This rule is often useful when one function is a power of x and the other function is a trigonometric function or e raised to a power of x.

Note that it may be necessary to repeat the integration by parts several times, depending on the situation.

Proving Integration by Parts

The proof for Integration by Parts is simple and important. Given:


It is known from the Product rule that:

(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)

If both sides are integrated:

f(x)g(x) = \int f'(x)g(x)dx + \int f(x)g'(x)dx

By rearranging the terms:

\int f(x)g'(x)dx = f(x)g(x) - \int f'(x)g(x)dx

Sometimes, Integration by Parts is simply expressed as:

\int udv = uv - \int vdu

Using Integration by Parts

Example 1

A common example where integration by parts is needed is this:

\int x\sin(x)dx

Let's set:

u = x
dv = sin(x)

We can begin to solve:

du = dx
v = − cos(x)
\int x\sin(x)dx = -x\cos(x) - \int -\cos(x) dx
= -x\cos(x) + \int \cos(x) dx

Now we are left with an integral that we do know how to solve:

= − xcos(x) + sin(x) + C


\int x\sin(x)dx = -x\cos(x) + sin(x) + C

Example 2

Say that we are given:

\int e^x \sin(x)dx

Let's set:

u = sin(x)
dv = exdx

And continue solving:

du = cos(x)dx
v = ex
\int e^x \sin(x)dx

And continue solving:

= \sin(x)e^x - \int e^x \cos(x) dx

Now, we set:

u = cos(x)
dv = exdx

And continue solving:

du = − sin(x)dx
v = ex
\sin(x)e^x - \int e^x \cos(x) dx
= \sin(x)e^x - \cos(x)e^x + \int -e^x\sin(x) dx = \sin(x)e^x - \cos(x)e^x - \int e^x\sin(x) dx

Let's look back at what we have done:

\int e^x \sin(x)dx = \sin(x)e^x - \cos(x)e^x - \int e^x\sin(x) dx

The integral of the function has the integral of itself ... inside itself. This is actually a common occurrence with functions like the one we are solving. However, we can cheat the never-ending cycle of integrals by doing this:

2\int e^x \sin(x)dx = \sin(x)e^x - \cos(x)e^x
\int e^x \sin(x)dx = {\sin(x)e^x - \cos(x)e^x \over 2}

Rapid Repeated Integration

Rapid Repeated Integration is a shortcut method for reduction problems that require Integration by Parts. It is especially useful when one function's derivative reduces to zero.

Example 3

For example, integrating:

\int x^4 sin(x)\,dx

We start off by making a table of x4 and sin(x). On the first column, the derivatives of x4 are taken until they reach zero. In the second column, sin(x) is integrated once down each row:

f(x) g(x)
x4 sin(x)
4x3 − cos(x)
12x2 − sin(x)
24x cos(x)
24 sin(x)
0 − cos(x)

Terms are multiplied diagonally from the left to the right and we add the product to the product of the next product, alternating signs with each step.

The first term, for example, is:

(x4)( − cos(x)) = − x4cos(x)

So then the integral becomes:

\int x^4 sin(x)\,dx = \ -x^4cos(x)-(-4x^3sin(x))+(12x^4cos(x))-(24xsin(x))-(-24cos(x))
=\ -x^4cos(x)+4x^3sin(x)+12x^2cos(x)-24xsin(x)-24cos(x)+C

While this method is very useful when part of the integrand becomes zero through derivation, it does not work in cases such as Example 2 above.

See Also


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