If U and V Are Functions of X, the Product Rule for Differentiation That We Met Earlier

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Ifuandvare functions ofx, theproduct rule for differentiationthat we met earlier gives us:

Rearranging, we have:

Integrating throughout, with respect tox, we obtain the formula forintegration by parts:

This formula allows us to turn a complicated integral into more simple ones. We must make sure we chooseuanddvcarefully.

Functionuis chosen so thatissimplerthanu.

Priorities for Choosingu

1. Letu= lnx

2. Letu=xn

3. Letu = enx

Example 1:

Solution:

We could letu=xoru= sin 2x.In general, we choose the one that allowsto be of a simpler form thanu.

So for this example, we chooseu=xanddv= sin 2x dx.

Substituting into the integration by parts formula, we get:

We could letu=xor.

Once again, we choose the one that allowsto be of a simpler form thanu, so we chooseu=x.

Thereforedu = dx. It follows thatdvmust be given by:.

Substituting into the integration by parts formula, we get:

Example 3:

Answer

We could letu=x2or.

Considering theprioritiesgiven above, we chooseu= ln 4xand so we must letdv =x2dx.

Substituting, we get:

Example 5:

Answer

Now, the integral we are left with cannot be found immediately. We need to perform integration by parts again, for this integral.

So putting this answer together with the answer for the first part, we have the final solution:

Example 6:

Answer

In this section, we see how to integrate expressions like

We substitute the following to simplify the expressions to be integrated:

Example 1:

Answer

We need to use:

witha= 3

We simplify the denominator of the question before proceeding to integrate:

(Some explanation for what we just did:)

93/2= (√9)3= 33= 27

tan2θ + 1= sec2θ

(a2)3/2=a3

Now, substituting

and

into the given integral gives us:

We now need to get our answer in terms ofx(since the question was in terms ofx).

Since we letx= 3 tan θ, we get

and we can draw a triangle to find the value of sin θ :

Hence, we notice that

Therefore, we can conclude that:

Top of Form

Bottom of Form

Example 2:.

Answer

Letx= 4 sec θ thendx =4 sec θ tan θdθ and
x2=16 sec2θ

Simplifying the square root part:

Substitutingdx =4 sec θ tan θdq,x2= 16 sec2xandinto the given integral gives us: [taking indefinite case first]

Since we letx= 4 sec θ , we get

Using a triangle, we can also derive that:

and

Therefore, we can conclude that:

1.

Answer

Letx= 4 sin θ , sodx= 4 cos θdθ

So

2.

Answer

Letx= 2 sin θ , sodx= 2 cos θdθ

3.

Answer

Firstly, note that

If we putu=x+ 1, thendu = dxand our integral becomes:

Now, we useu= sec θ and so:du =sec θ tan θdθ

Returning to our integral, we have:

Example 1:Integrate:

Answer

We could either chooseu= sinx,u= sin1/3xoru= cosx. However, only the first one of these works in this problem.

So we let

u= sinx.

Finding the differential:

du= cosx dx

Substituting these into the integral gives:

Example 2:Integrate:

Answer

We have some choices foruin this example:sin-14x,1 − 16x2,or√(1 − 16x2).Only one of these gives a result forduthat we can use to integrate the given expression, and that's the first one.

So we letu =sin-14x.

Then, using thederivative of the inverse sine, we have:

We divide both sides by 4 so we can substitue into our original expression:

Now to complete the required subsitution (u =sin-14xand thedu/4 expression we just found):

The expression on the right is a simple integral:

To complete the porblem, we substitutesin-14xforu: