Thursday, May 3, 2012

Calculus Revisited #12: Integration by Parts

Welcome to Part 12 of our 21-part Calculus Revisited series. Today we will cover Integration by Parts in our second of four integration techniques.

Integration by Parts

∫ u(x) * v'(x) dx = u(x) * v(x) - ∫ v(x) * u'(x) dx

Pick the parts u(x) and v'(x) dx. Find the derivative of u(x) and the integral of v'(x).

The trick is to pick the parts u(x) and v'(x) so that v(x) * u'(x) is easier to integrate than u(x) * v'(x).

In practice, you may find that somewhere down the line, u(x) * v'(x) = v(x) * u'(x). This comes up especially where you have sines and cosines.

You may have to repeat this procedure more than once. Let's get to some examples.

Problems

The key to this technique is to pick the correct parts. Again, practice, practice, practice.

If you can pick u(x) such that du/dx is a constant, or goes towards a constant, it will most likely help.

Indefinite Integrals

1. ∫ x sin x dx

u = x
dv = sin x dx

du = 1 dx (take the derivative of u(x))
v = - cos x (take the integral of v(x))

Putting it together:

∫ x sin x dx
= (x) * ( - cos x ) - ∫ (- cos x) (1) dx
= - x * cos x + ∫ cos x dx

The second term is easy to integrate.

= -x * cos x + sin x

For indefinite integrals, you can tack on the "C" at the end

Final Answer: ∫ x sin x dx = -x* cos x + sin x + C

2. ∫ x * 3^x dx

u = x
dv = 3^x dx

du = 1 dx
v = 3^x / ln 3

Then:

∫ x * 3^x dx
= ( x ) * ( 3^x / ln 3 ) - ∫ ( 3^x / ln 3) * 1 * dx
= ( x * 3^x / ln 3 ) - ( 1 / ln 3 ) * ∫ 3^x dx

Again, v(x) * u'(x) is easier to integrate than u(x) * v'(x), which is the idea.

= ( x * 3^x / ln 3 ) - ( 1 / ln 3) * ( 3^x / ln 3 )
= ( 3^x / ln 3) * (x - 1 / ln 3 )

Final Answer: ∫ x * 3^x dx = ( 3^x / ln 3) * (x - 1 / ln 3 ) + C

In the next two problems, integration by parts is used twice.

3. ∫ x^2 * e^x dx

Let's see if we can simplify that x^2 a bit..

u = x^2
dv = e^x dx

du = 2x dx
v = e^x

∫ x^2 * e^x dx
= x^2 * e^x - ∫ 2x* e^x dx

Looks like we have to use integration by parts again.

Concentrate on ∫ 2x * e^x dx:

u = 2x
dv = e^x dx

du = 2 dx
v = e^x

Then:
x^2 * e^x - ∫ 2x* e^x dx
= x^2 * e^x - ( 2x * e^x - ∫ 2 * e^x dx )
= x^2 * e^x - 2x * e^x + 2 ∫ e^x dx

The third term is straight-forward

= x^2 * e^x - 2x * e^x + 2 * e^x
= e^x * (x^2 - 2x + 2)

Final Answer: ∫ x^2 * e^x dx = e^x * (x^2 - 2x + 2) + C

4. ∫ e^x * sin x dx

u = e^x
dv = sin x dx

du = e^x dx
v = -cos x

∫ e^x * sin x
= -e^x * cos x - ∫-e^x * cos x dx
= -e^x * cos x + ∫e^x * cos dx

Let's do the technique again.

Concentrate on ∫ e^x * cos dx

u = e^x
dv = cos x dx

du = e^x dx
v = sin x

Then:
-e^x * cos x + ∫e^x * cos dx
= -e^x * cos x + (e^x * sin x - ∫e^x * sin x dx)

Note that:
∫ e^x * sin x dx = -e^x * cos x + e^x * sin x - ∫e^x * sin x dx

Combine the "∫e^x * sin x dx" terms, and solve for it.

2 * ( ∫ e^x * sin x dx ) = -e^x * cos x + e^x * sin x
∫ e^x * sin x dx = 1/2 * (-e^x * cos x + e^x * sin x)

Final Answer:
∫ e^x * sin x dx = e^x / 2 * (-cos x + sin x) + C


Definite Integrals

5. Calculate

3
∫ x^2 * ln x dx
1

First find the antiderivative F(x) - then we can worry about using the limits later. This is one way to do it.

F(x) = ∫ f(x) dx

We know

b
∫ f(x) dx = F(b) - F(a)
a

F(x) = ∫ x^2 * ln x dx

u = ln x
dv = x^2 dx

du = 1/x dx
v = x^3 / 3

Then:

F(x)
= x^3 / 3 * ln x - ∫ 1/x * x^3 /3 dx
= x^3 * ln x * 1/3 - ∫ x^2 / 3 dx
= 1/3 * x^3 *ln x - x^3 / 9

Now:

3
∫ x^2 ln x dx = F(3) - F(1)
1

= ( (1/3 * 3^3 * ln 3) - (3^3 / 9) ) - ( (1/1 * 1^3 * ln 1) - (1^3 / 9) )
= ( 9 * ln 3 - 3 ) - ( - 1/9 )
= 9 * ln 3 - 3 + 1/9
= 9 * ln 3 - 26/9 ≈ 6.9982

Next time we will be working with integrals involving trigonometric identities.

Until then, take care!

Eddie

This blog is property of Edward Shore. © 2012



1 comment:

  1. This is good. But, how can we select which function is to be taken first(u) and which one is to be taken second(v)? You didn't mention any idea about it, Otherwise it is really good. This problem can be solved by introducing "ILATE" rule of integration,Which tells us that use the Inverse function first then Logarithmic function then Algebraic function then Trigonometric function and then Exponential function, while performing integration by parts.I hope you will find my suggestion correct.

    ReplyDelete