Integration - Indefinite - By Parts and U-Sub

[FaraDazed]

Homework Statement

Integrate the following indefinite integrals

A:$\int e^x (x^2+1) dx$

B:$\int e^x cos(3x+2) dx$

Homework Equations

$\int u dv = uv - \int v du$

The Attempt at a Solution

Part A: I have done the following but when I use an integration calculator online its not what I have (although I suppose there's a few ways of doing it)

$$let \, u=x^2 +1 \,\, ∴ \, du=2x \\<br /> let \, dv=e^x \,\, ∴ \, v=e^x \\<br /> \\<br /> ∴e^x(x^2+1)-\int e^x 2x \, dx \\$$
Then doing another by parts on that integral
$$let \, u=2x \,\, ∴ \, du=2 \\<br /> let \, dv=e^x \,\, ∴ \, v=e^x \\<br /> \\<br /> 2xe^x-2\int e^x \, dx = 2xe^x-2e^x + C$$
Then plugging that into the first bit I get
$$e^x(x^2+1)-2xe^x-2e^x + C$$For Part B:
I am really confused as it doesn't look like a u-sub problem and no matter which way around I set the variables to do a by parts problem, the resulting integral is no easier to solve than the original so any hints on where to start with that one would be much appreciated.

Thanks :)

 

[PeroK]

For part a), you've been a bit sloppy: omitted the dx on the integral, omitted the constant of integration and made a simple mistake on the second parts. I'd try again and be more careful.

For part b): maybe things will work out if you just keep going.

 

[FaraDazed]

For part a), you've been a bit sloppy: omitted the dx on the integral, omitted the constant of integration and made a simple mistake on the second parts. I'd try again and be more careful.

For part b): maybe things will work out if you just keep going.

I can assure you that whilst I did my problem on paper, and in doing so did not forget the dx's and constant, when I came to type it into here using latex it was something I forgot to add yes sorry. I have now corrected the sloppyness, I am very new to integration by parts and well integration as a whole really. I would appreciate it if you could let me know where my error is on the second bit as I have gone through it several times and I might be blind to it by now. Thanks :)

 

[PeroK]

Watch your pluses and minuses!

 

[HallsofIvy]

Homework Statement

Integrate the following indefinite integrals

A:$\int e^x (x^2+1) dx$

B:$\int e^x cos(3x+2) dx$

Homework Equations

$\int u dv = uv - \int v du$

The Attempt at a Solution

Part A: I have done the following but when I use an integration calculator online its not what I have (although I suppose there's a few ways of doing it)

$$let \, u=x^2 +1 \,\, ∴ \, du=2x \\<br /> let \, dv=e^x \,\, ∴ \, v=e^x \\<br /> \\<br /> ∴e^x(x^2+1)-\int e^x 2x \, dx \\$$
Then doing another by parts on that integral
$$let \, u=2x \,\, ∴ \, du=2 \\<br /> let \, dv=e^x \,\, ∴ \, v=e^x \\<br /> \\<br /> 2xe^x-2\int e^x \, dx = 2xe^x-2e^x + C$$
Then plugging that into the first bit I get
$$e^x(x^2+1)-2xe^x-2e^x + C$$

You have a sign error here. (-)(-)= +.

For Part B:
I am really confused as it doesn't look like a u-sub problem and no matter which way around I set the variables to do a by parts problem, the resulting integral is no easier to solve than the original so any hints on where to start with that one would be much appreciated.

Thanks :)

Yes, use integration by parts.
To start, let [itex]u= e^x[itex], $dv= sin(3x+ 2)dx$ so that $du= e^xdx$ and $v= -(1/3)cos(3x+ 2)$. <br /> $$\int e^xsin(3x+ 2)dx= -(1/3)e^xcos(3x+2)+ (1/3)\int e^x cos(3x+ 2) dx$$<br /> <br /> Now do pretty much the same thing: let $u= e^x$, $dv= cos(3x+ 2)dx$ so that $du= e^xdx$ and $v= (1/3)sin(3x+ 2)$. That gives<br /> $$\int e^x sin(3x+ 2)dx= -(1/3)e^x cos(3x+2)+ (1/3)[(1/3)e^xsin(3x+2)- (1/3)\int e^x sin(3x+ 2)dx$$<br /> $$\int e^x sin(3x+ 2)dx= -(1/3)e^x cos(3x+2)+ (1/9)e^xsin(3x+2)- (1/9)\int e^x sin(3x+ 2)dx$$<br /> <br /> Now, <b>don't</b> try to integrate that last integral on the right- it is the same as on the left. Just add $(1/9)\int e^x sin(3x+ 2)dx$ to both sides and divide by 2.[/itex][/itex]

 

[FaraDazed]

Watch your pluses and minuses!

Ah right, yeah now after checking one of the integration calculators I can see that there should be a +2e^x term in it whereas I have a -2e^x term but still going through my calculation step by step I can't see how it becomes positive. I am sure there must be a negative times a negative somewhere but I cannot see it. It may be something to do with my understanding on how you do integration by parts when you have to do it twice by parts.

 

[jaytech]

For part 2, my personal preference is to substitute for everything in the cos() term. So for me I'd just say 3x+2 = s and then you have 3dx = ds and the e^x becomes e^(s-2)/3. Then when I have a solution as a function of s I plug 3x+2 back in. I just think that doing the by parts with multiple variables in the exponent is easier. Just my preference.

 

[PeroK]

$$2xe^x-2\int e^x \, dx = 2xe^x-2e^x + C$$
Then plugging that into the first bit I get
$$e^x(x^2+1)-2xe^x-2e^x + C$$

Your mistake is nothing to do with parts, it's to do with -(-2) = 2:

$$e^x(x^2+1)-(2xe^x-2e^x) + C = e^x(x^2+1)-2xe^x+2e^x + C$$

 

[FaraDazed]

Your mistake is nothing to do with parts, it's to do with -(-2) = 2:

$$e^x(x^2+1)-(2xe^x-2e^x) + C = e^x(x^2+1)-2xe^x+2e^x + C$$

Ah right yeah I can see it now. Sounds stupid but I couldn't see it as when I scan it and visualized it in my head, I was visualizing the end result with the parentheses around it like (2xe^x-2e^x) and therefore couldn't see as my answer is correct if I had the parentheses around that bit.I can see now though (thanks to you :) ) that that means without them I have to flip the signs.

Thanks :) I am off to bed for the night and have the next two days to go through part B. Thanks.