اعداد : كارزان خالد /

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.

u = x / dv= sin 2x dx
du = 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:.

u = x /
du = dx /

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.

u =ln 4x / dv =x2dx

Substituting, we get:

Example 5:

Answer

u=x2 / dv = e-xdx
du =2x dx / v = -e-x

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

u=x / dv = e-xdx
du = dx / v = -e-x

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

Example 6:

Answer

u= lnx / dv = dx
/ v = x

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:

/ NOTE: We could have changed upper and lower limits for θ here, and there would be no need to convert our expression back in terms ofx.

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: