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.

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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

Let

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

=(u^4)/(4)+K

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

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

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

Let

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

=2((2)/(3))u^[3/2]+K

=4/3(1-e^-x)^[3/2]+K

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

=(u^3)/(3)+K

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

When x=1, y=2, so:

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

K=2

Therefore the equation is

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

Continues below

Exercises

Integrate each of the following functions:

Exercise 1

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

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):

(d(cos^-1u))/(dx)=(-1)/sqrt(1-u^2)(du)/(dx)

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

So

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[(u^2)/(2)]_[x=1]^[x=e]

=[(-u^2)/(4)]_[x=1]^[x=e]

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

=[(-(1-2(1))^2)/(4)]-[(-(1-0)^2)/(4)]

=0

Exercise 3

int(e^x+e^(-x))^(1/4)(e^x-e^(-x))dx

int(e^x+e^-x)^(1/4)(e^x-e^-x)dx

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

Then

int(e^x+e^-x)^(1/4)(e^x-e^-x)dx=intu^(1/4)du

=(4u^(5//4))/5+K

=(4(e^x+e^-x)^(5//4))/5+K

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

So

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

=-int_(theta=pi//3)^(theta=pi//2)u^(-1//2)du

=-2[u^(1//2)]_(theta=pi//3)^(theta=pi//2)

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

=-2[sqrt(1+0)-sqrt(1+0.5)]

=0.449489

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

=1/2intu^2du

=(u^3)/6+K

=((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

1=1.674539+K

This gives:

K~~-0.675

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

v=[ln^2(t^3+1)](t^2)/(t^3+1)

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:

Now

s=int ("velocity") dt

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

We need to find:

s=int[ln^2(t^3+1)](t^2)/(t^3+1)dt

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:

(du)/(dt)=(du)/(dv)(dv)/(dt)

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

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

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)

So

s=1/3int[u^2]du

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

=u^3/9+K

=(ln^3(t^3+1))/9+K

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

s=(ln^3(t^3+1))/9

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

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