What gives the integration of cosine

Partial integration

Repetition: product rule as derivation rule

The partial integration is a technique for integrating special functions. These functions consist of the product of two functions, their Derivatives are known.

In order to better understand the derivation of the partial integration, you have to refer to the Product rule recall. This is one Derivation rule. We use it to derive the product of functions $ u (x) \ cdot v (x) $:

$ \ left (u (x) \ times v (x) \ right) '= u' (x) \ times v (x) + u (x) \ times v '(x) $.

In your Formula collection you might also find the abbreviation $ (uv) '= u'v + uv' $.

Derivation of the partial integration

With the help of Product rule can the partial integration can be derived: The product rule says that two functions are equal: $ \ left (u (x) \ cdot v (x) \ right) '$ is the same as $ u' (x) \ cdot v (x) + u ( x) \ cdot v '(x) $.

So they are also correct indefinite integrals these functions match. We see this by integrating both sides after the variable. ("We just put an integral in front of it on both sides.")

$ \ begin {array} {lll} \ int \ left (u (x) \ cdot v (x) \ right) 'dx & = & \ int \ left (u' (x) \ cdot v (x) + u (x) \ cdot v '(x) \ right) dx \ & = & \ int \ left (u' (x) \ cdot v (x) \ right) dx + \ int \ left (u (x) \ cdot v '(x) \ right) dx \ end {array} $

Since the Differentiation and the integration cancel each other out, i.e. $ \ int f '(x) dx = f (x) $, we can transform it as follows:

$ u (x) \ times v (x) = \ int \ left (u '(x) \ times v (x) \ right) dx + \ int \ left (u (x) \ times v' (x) \ right ) dx $

Now we subtract the term $ \ int \ left (u (x) \ cdot v '(x) \ right) dx $ on both sides of the equation:

$ u (x) \ cdot v (x) - \ int \ left (u (x) \ cdot v '(x) \ right) dx = \ int \ left (u' (x) \ cdot v (x) \ right) dx $

For a better overview, let's swap the two sides of the equation:

$ \ int \ left (u '(x) \ cdot v (x) \ right) dx = u (x) \ cdot v (x) - \ int \ left (u (x) \ cdot v' (x) \ right) dx $

The rule shown here is called partial integration or Product integration. The following abbreviated form is easier to remember:

$ \ int (u'v) dx = uv- \ int (uv ') dx $

When do you use partial integration?

The partial integration we apply to integrate a function that consists of two or more factors. Examples of such functions are:

  • $ f (x) = x \ cdot e ^ x $
  • $ g (x) = \ sin (x) \ cdot \ cos (x) $
  • $ h (x) = x \ cdot \ ln (x) $

What should you watch out for with partial integration?

You can see that on both sides of the partial integration is an integral. What should be an advantage here? We want it to be easier and not harder!

For this purpose, when integrating a product function, we have to consider which of the two factors “plays the role of $ u '(x) $”. From this function we also need the Indefinite integral know. The other function "then plays the role of $ v (x) $".

example

Let us consider the function $ f (x) = x ^ 2 \ cdot x $. Of course, you can also integrate this function more easily using the power rule of integration. Namely, $ f (x) = x ^ 2 \ cdot x = x ^ 3 $ and thus:

$ \ int f (x) dx = \ frac14x ^ 4 + c $

However, this example is well suited to explain the technique of partial integration and to illustrate its validity. For the partial integration we now assume the following functions:

We also need the derivative of $ v (x) $ and the function $ u (x) $:

  • $ u (x) = \ frac 13 x ^ 3 $
  • $ v '(x) = 1 $

This results in the following calculation:

$ \ begin {array} {lll} \ int \ underbrace {x ^ 2} _ {u '(x)} \ cdot \ underbrace {x} _ {v (x)} dx & = & \ underbrace {\ frac13x ^ 3} _ {u (x)} \ cdot \ underbrace {x} _ {v (x)} - \ int \ underbrace {\ frac13x ^ 3} _ {u (x)} \ cdot \ underbrace {1} _ { v '(x)} dx \ & = & \ frac13x ^ 4- \ frac1 {12} x ^ 4 + c \ & = & \ frac3 {12} x ^ 4 + c \ & = & \ frac14 x ^ 4 + c \ end {array} $

This antiderivative with $ c \ in \ mathbb {R} $ corresponds to the antiderivative above, which we determined using the power rule.

Exercises for the application of partial integration

In the following we take a look at some examples, which of the two factors of the product function to be integrated should play the role of $ u '(x) $ and how we partial integration finally apply.

Exponential functions

We want to calculate the indefinite integral $ \ int (x \ cdot e ^ x) dx $.

We choose $ u '(x) = e ^ x $ and $ v (x) = x $ because the antiderivative of $ u' (x) $, namely $ u (x) = e ^ x $, is known and because $ v '(x) = 1 $ is easy to use.

At the Derive For polynomials, the exponent is always $ 1 $ smaller. That's why you vote for Exponential functions of the above form always the polynomial as $ v (x) $ and the exponential factor as $ u '(x) $. This is often called a Clearing polynomials designated.

Now we can partial integration apply.

$ \ begin {array} {lll} \ int (x \ cdot e ^ x) dx & = & x \ cdot e ^ x- \ int (1 \ cdot e ^ x) dx \ & = & x \ cdot e ^ xe ^ x + c \ & = & (x-1) \ cdot e ^ x + c \ end {array} $

Undoubtedly the right integral is easier to calculate. If we want to integrate $ f (x) = x ^ 2 \ cdot e ^ x $, we proceed in the same way. However, it has to be partially integrated twice.

Trigonometric functions

A common area of ​​application for the partial integration are the trigonometric functions. Let's take as example the function $ f (x) = \ sin (x) \ cdot \ cos (x) $. Let us first determine what we will need later:

  • $ u '(x) = \ sin (x) $
  • $ u (x) = - \ cos (x) $
  • $ v (x) = \ cos (x) $
  • $ v '(x) = - \ sin (x) $

Then we can start calculating.

$ \ begin {array} {lll} \ int (\ sin (x) \ cdot \ cos (x)) dx & = & - \ cos (x) \ cdot \ cos (x) - \ int ((- \ cos (x)) \ cdot (- \ sin (x))) dx \ & = & - (\ cos (x)) ^ 2- \ int (\ sin (x) \ cdot \ cos (x)) dx \ end {array} $

As we can see, the starting integral appears again. If we add $ \ int (\ sin (x) \ cdot \ cos (x)) dx $ on both sides, we get

$ \ begin {array} {lll} 2 \ int (\ sin (x) \ cdot \ cos (x)) dx & = & - (\ cos (x)) ^ 2 & | &: 2 \ \ int (\ sin (x) \ cdot \ cos (x)) dx & = & - \ frac12 (\ cos (x)) ^ 2 \ end {array} $

Logarithmic functions

Finally, we use a trick to integrate logarithmic functions. Let's consider the simple example $ f (x) = \ ln (x) $. To partially integrate this function, we add the factor $ 1 $, since we need a product: $ \ int \ ln (x) dx = \ int (1 \ cdot \ ln (x)) dx $. Then the following applies:

  • $ u '(x) = 1 $
  • $ u (x) = x $
  • $ v (x) = \ ln (x) $
  • $ v '(x) = \ frac 1x $

The partial integration then looks like this:

$ \ begin {array} {lll} \ int \ ln (x) dx & = & \ int (1 \ cdot \ ln (x)) dx \ & = & x \ cdot \ ln (x) - \ int \ left (x \ cdot \ frac1x \ right) dx \ & = & x \ cdot \ ln (x) - \ int 1dx \ & = & x \ cdot \ ln (x) -x + c \ end {array} $

Since the derivative of $ \ ln (x) $ is just $ \ left (\ ln (x) \ right) '= \ frac1x $, $ x $ and $ \ frac1 {x} $ cancel each other out. Therefore you always choose the polynomial as $ u '(x) $ and the logarithm as $ v (x) $.