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1. Integration: The General Power Formula

by M. Bourne

In this section, we apply the following formula to trigonometric, logarithmic and exponential functions:

`intu^ndu=(u^(n+1))/(n+1)+C\ \ \ (n!=-1)`

(We met this substitution formula in an earlier chapter: General Power Formula for Integration.)

Example 1: Integrate: `intsin^(1//3)\ x cos x dx`


`int sin^[1//3]x\ cos x\ dx`

Our options are to either choose u = sin x, u = sin1/3 x or u = cos x. However, only the first one of these works in this problem.

So we let

u = sin x.

Finding the differential:

du = cos x dx

Substituting these into the integral gives:

`int sin^[1//3]x\ cos x\ dx = int u^[1//3] du`

`= (3u^[4//3])/(4)+K`

`=(3\ sin^[4//3]\ x)/(4)+K`

The last line is obtained by re-expressing our answer in terms of x.

Example 2: Integrate: `int(sin^(-1)4x)/sqrt(1-16x^2)dx`


`int (sin^-1 4x)/(sqrt[1-16x^2]) dx`

We have some choices for u in this example. Either `sin^-1 4x`, or `1 − 16x^2`, or `sqrt(1 − 16x^2)`. Only one of these gives a result for du that we can use to integrate the given expression, and that's the first one.

So we let `u=sin^-1 4x`

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

`du=(4)/(sqrt[1-16x^2]) dx`

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

`1/4 du=(1)/(sqrt[1-16x^2]) dx`

Now to complete the required subsitution (u = sin-14x and the `(du)/4` expression we just found):

`int(sin^-1 4x)/(sqrt[1-16x^2]) dx=1/4intu\ du`

The expression on the right is a simple integral:

`1/4intu\ du = 1/4((u^2)/(2))+K`

To complete the problem, we substitute sin-14x for u:

`int(sin^-1 4x)/(sqrt[1-16x^2]) dx` ` = ((sin^-1 4x)^2)/(8)+K`

Example 3: Integrate: `int((3+ln\ 2x)^3)/xdx`


`int ((3+ ln\ 2x)^3)/(x) dx`


u = 3 + ln 2x

We can expand out the log term on the right hand side as follows:

3 + ln 2x = 3 + ln 2 + ln x

Now the first 2 terms on the right are constants (whose derivative equals zero) and the derivative of the natural log of x is `1/x`.

Then `du = 1/x dx`.

`int(3+ln\ 2x)^3/(x)dx =int u^3 du`


`=((3+ln\ 2x)^4)/(4)+K`

Example 4: Integrate: `int2sqrt(1-e^(-x))e^(-x)dx`


`int2 sqrt[1-e^[-x]]e^[-x] dx`


`u = 1 − e^(-x)`

The derivative of u is

`(du)/dx = 0 − (-e^(-x)) = e^(-x)`

So the differential du is:

`du = e^(-x)dx`

We substitute to give:

`int2 sqrt[1-e^-x] e^-x dx=2 int sqrtu\ du`

`=2int u^[1/2]\ du`



Example 5: Find the equation of the curve for which `(dy)/(dx)=((ln\ x)^2)/x` if the curve passes through `(1, 2)`.


`y=int((ln\ x)^2)/(x)\ dx`

Let `u= ln\ x`.

Then `du = 1/x dx`

`y=int((ln\ x)^2)/(x) dx= int u^2 du`


`= ((ln\ x)^3)/(3)+K`

When `x=1`, `y=2`, so:

`2=((ln\ text[l])^3)/(3)+K`


Therefore the equation is

`y=((ln\ x)^3)/(3)+2`


Integrate each of the following functions:

Exercise 1



`int((cos^-1 2x)^4)/(sqrt[1-4x^2]) dx`

Put `v = cos^-1 2x`

(since this is the only substitution that works. The other "likely" one, `v=1-4x^2`, doesn't give us anything useful when we differentiate while doing the integral. I'm using `v` this time, so as not to confuse things with `u` in the following formula.)

We need to find `(dv)/dx`.

In general, (from 3. Derivatives of the Inverse Trigonometric Functions):


In this example, `u=2x`, so we have `(du)/dx = 2`.

Thus `(dv)/(dx) = (d(cos^-1 2x))/(dx)= (dv)/(du) (du)/dx ` `= (-1)/sqrt(1-(2x)^2)(2) = (-2)/(sqrt[1-4x^2])`

Now our integral doesn't have `-2` as a constant anywhere, but it does have `1/(sqrt[1-4x^2])dx`, so we'll write our differential expression as follows, by dividing throughout by `-2`:

`-1/2dv=1/(sqrt[1-4x^2]) dx`


`int((cos^-1 2x)^4)/(sqrt[1-4x^2]) dx = - 1/2 int v^4 dv`

`=(-1/2) ((v^5)/(5))+K`

`=(-(cos^-1 2x)^5)/(10)+K`

Exercise 2

`int_1^e((1-2 ln x))/xdx`


`int_1^e ((1-2\ ln\ x))/(x) dx`

Put `u = 1 - 2\ ln\ x`, then `du=(-2)/(x) dx`

`int_1^e ((1-2\ ln\ x))/(x) dx = -1/2 int_[x=1]^[x=e] u\ du`



`=[(-(1-2\ ln\ x)^2)/(4)]_1^e`



Exercise 3




Put `u=e^x+e^-x`, so `du=(e^x-e^-x)dx`





Exercise 4

`int_(pi//3)^(pi//2)(sin theta\ d theta)/(sqrt(1+cos theta)`


`int_(pi//3)^(pi//2)(sin theta\ d theta)/(sqrt(1+cos theta`

Put `u=1+cos theta`, then `du=-sin theta\ d theta`


`int_(pi//3)^(pi//2)(sin theta\ d theta)/(sqrt(1+cos theta))=-int_(theta=pi//3)^(theta=pi//2)(du)/sqrtu`



`=-2[sqrt(1+cos theta)]_(pi//3)^(pi//2)`



Exercise 5

Find the equation of the curve for which `(dy)/(dx)=(1+tan 2x)^2sec^2 2x` if the curve passes through `(2, 1)`.


`(dy)/(dx)=(1+tan x)^2sec^2 2x`

We need to find:

`y=int(1+tan 2x)^2sec^2\ 2x\ dx`

Put `u=1+tan 2x`, then `du=2\ sec^2 2x\ dx`

`y = int(1+tan 2x)^2 sec^2 2x\ dx`



`=((1+tan 2x)^3)/6+K`

The curve passes through `(2, 1)`.

This means when `x = 2`, `y = 1`. Substituting gives:

`1=((1+tan 2(2))^3)/6+K`

`1=((1+tan 4)^3)/6+K`


This gives:


So we finally have the required equation for y:

`y=((1+tan 2x)^3)/6-0.675`

Exercise 6

A space vehicle is launched vertically from the ground such that its velocity v (in km/s) is given by


where t is the time in seconds. Find the altitude of the vehicle after 10.0 s.

The graph of `v=[ln^2(t^3+1)](t^2)/(t^3+1)` is as follows:



`s=int ("velocity") dt`

[We could use s or h for height in this problem.]

We need to find:


We put the following (since any other possible substitution doesn't actually work):

`u = ln(t^3+ 1)`

We need to find `(du)/(dt)`. We have a function of a function, so we'll need to use the Chain Rule to find that derivative. We substitute a new variable, `v`:

`v=t^3+1`, giving `u = ln v`.

Now `(dv)/(dt) = 3t^2` and `(du)/(dv) = 1/v = 1/(t^3+1)`.

Using the Chain Rule, we get:


` = 1/(t^3+1) xx (3t^2)`


Writing this as a differential in the form we need for the integral, we have:

`1/3 du=(t^2)/(t^3+1)dt`

Going back to the original integral with the substitution `u = ln(t^3+ 1)`, we get:

`u^2 = ln^2(t^3+ 1)`



`=1/3 xx u^3/3+K`



Now, since the height is `0` when `t = 0`, we substitute and obtain `K = 0`.


At `t=10`, the height of the space vehicle will be:

`s=(ln^3((10)^3+1))/9=36.6\ "km"`

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