Integrating Trigonometric and Exponential Functions with Integration by Parts

[MHrtz]

Homework Statement

Take the integral of the following:

1. ln(2x+1)
2. arctan4x
3. e^cosxsin2x evaluated from 0 to pi

The Attempt at a Solution

1. took the derivative of ln(2x+1) and integrated dx. my solution was:

xln(2x +1) + x + [(2x + 1)^-2]/2 + C

The books answer was:

(1/2)(2x + 1)ln(2x + 1) - x + C

2. took the derivative of arctan4x and the integrated dx. My solution was:

xarctan4x - (1+ x²)^-2 + C

The books answer was:

xarctan 4x - (1/8)ln(1 + 16x²) + C

3. I derived sin2x and integrated e^cosx but I got something more complex than what I started with.

 

[Mark44]

Homework Statement

Take the integral of the following:

1. ln(2x+1)
2. arctan4x
3. e^cosxsin2x evaluated from 0 to pi

The Attempt at a Solution

1. took the derivative of ln(2x+1) and integrated dx. my solution was:

xln(2x +1) + x + [(2x + 1)^-2]/2 + C

I'm not sure that you understand how integration by parts works. When you do this, you get a product of functions (xln(2x + 1) in this case) and another integral, that should be simpler than the one you started with.

\int \ln{(2x+1)}dx = x~ln(2x + 1) - \int \frac{2xdx}{2x + 1}
You didn't mention getting the integral on the right. If you did get it, I don't think you integrated correctly.

The books answer was:

(1/2)(2x + 1)ln(2x + 1) - x + C

2. took the derivative of arctan4x and the integrated dx. My solution was:

xarctan4x - (1+ x²)^-2 + C

The books answer was:

xarctan 4x - (1/8)ln(1 + 16x²) + C

3. I derived sin2x and integrated e^cosx but I got something more complex than what I started with.

 

[MHrtz]

I did get the integral on the right then I made u = 2x + 1, du = 2 dx, and (u - 1) = 2x

 

[Char. Limit]

For the third integral, start by changing sin(2x) to 2 sin(x) cos(x) using the double angle rule, then substitute u=cos(x). Only then should you integrate by parts.

 

[MHrtz]

If u = cosx then does this mean that dv = 2sinx e^cosx dx?

 

[Char. Limit]

No, I mean make a u-substitution first. THEN integrate by parts. Use s=cos(x) if you're too used to using u and dv for integration by parts.

 

[Mark44]

I did get the integral on the right then I made u = 2x + 1, du = 2 dx, and (u - 1) = 2x

So it looks like you are doing an ordinary substitution. There are two other approaches that are more direct.
1. Rewrite the integrand of the integral on the right like so.
\frac{2x}{2x + 1} = \frac{2x + 1 - 1}{2x + 1} = \frac{2x + 1}{2x + 1} - \frac{1}{2x + 1} = 1 - \frac{1}{2x + 1}
So
-\int \frac{2xdx}{2x+1} = -\int\left(1 - \frac{1}{2x + 1} \right)dx
This is an easy integral.
2. Use polynomial division to divide 2x by 2x + 1. This results in 1 - 1/(2x + 1), the same as above.

 

[Mark44]

For the third integral, start by changing sin(2x) to 2 sin(x) cos(x) using the double angle rule, then substitute u=cos(x). Only then should you integrate by parts.

If u = cosx then does this mean that dv = 2sinx e^cosx dx?

Yes. When you integrate dv you will get -e^cosx. Integration by parts results in the following:
-cosx~e^{cosx} - \int e^{cosx}~sinx~dx

The integral on the right can be done with a substitution.

 

[Char. Limit]

Oh right... I guess you could do integration by parts that way too.

 

[MHrtz]

I tried your method for the third integral but still got a wrong answer. I used a calculator and determined that the actual answer was e - (1/e). Here's what I did:

s = cosx
ds = -sinx dx

\int2sinxcosxe^cosxdx = -2\intse^sds

u = s
du = ds
dv = e^sds
v = e^s

-2se^s - \inte^sds

-2cosxe^cosx + \intsinxe^cosx dx

Let me know if that much is right so I can continue.

 

[Mark44]

I tried your method for the third integral but still got a wrong answer. I used a calculator and determined that the actual answer was e - (1/e). Here's what I did:

s = cosx
ds = -sinx dx

\int2sinxcosxe^cosxdx = -2\intse^sds

u = s
du = ds
dv = e^sds
v = e^s

-2se^s - \inte^sds

-2cosxe^cosx + \intsinxe^cosx dx

In the line above, the + in the middle should be =. Also, there should be a coefficient of 2 on the integral at the right. These two errors would cause you to get an incorrect answer.

Let me know if that much is right so I can continue.

 

[MHrtz]

-2cosxe^cosx - 2\intsinxe^cosx dx

u = cosx
du = -sinx dx

2\inte^udu

-2cosxe^cosx + 2e^u

[-2cosxe^cosx + 2e^cosx] evaulated from 0 to pi

What about this?

 

[Mark44]

I get 4/e, which is the same as what wolframalpha gets. The value you reported earlier (e - 1/e) is apparently incorrect.

 

[MHrtz]

Ok. So I get 4/e now as my final answer. My mistake was that I dropped a negative while carrying out my evaluation. For the first integral I tried a different attempt but the answer is still off:

u = ln(2x+1)
du = 2/(2x+1) dx
dv = dx
v = x

xln(2x+1) - \int 2x/(2x+1) dx

u = 2x + 1
du = 2 dx
x = (u-1)

xln(2x+1) - (1/2)\int(u-1)/u du

xln(2x+1) - (1/2)\int1 - u^-1 du

xln(2x+1) - (1/2)x - (1/2)ln(2x + 1) + C

Can this be simplified to look like the book's answer or am I still integrating incorrectly?

 

[Mark44]

Ok. So I get 4/e now as my final answer. My mistake was that I dropped a negative while carrying out my evaluation. For the first integral I tried a different attempt but the answer is still off:

u = ln(2x+1)
du = 2/(2x+1) dx
dv = dx
v = x

xln(2x+1) - \int 2x/(2x+1) dx

u = 2x + 1
du = 2 dx
x = (u-1)

The line above is wrong. u = 2x + 1 ==> x = (u - 1)/2

xln(2x+1) - (1/2)\int(u-1)/u du

xln(2x+1) - (1/2)\int1 - u^-1 du

xln(2x+1) - (1/2)x - (1/2)ln(2x + 1) + C

Can this be simplified to look like the book's answer or am I still integrating incorrectly?

No and yes. See above.

 

[Char. Limit]

Ok. So I get 4/e now as my final answer. My mistake was that I dropped a negative while carrying out my evaluation. For the first integral I tried a different attempt but the answer is still off:

u = ln(2x+1)
du = 2/(2x+1) dx
dv = dx
v = x

xln(2x+1) - \int 2x/(2x+1) dx

u = 2x + 1
du = 2 dx
x = (u-1)

It's actually 2x = (u - 1). Which still gives the correct integral in the next line, I believe.

x ln(2x+1) - \frac{1}{2}\int\frac{u-1}{u} du

xln(2x+1) - (1/2)\int1 - u^-1 du

xln(2x+1) - (1/2)x - (1/2)ln(2x + 1) + C

Can this be simplified to look like the book's answer or am I still integrating incorrectly?