MHB How can I determine which integration technique to use for a specific problem?

[ineedhelpnow]

(Wave)I have a test tomorrow on the different Techniques of Integration: integration by parts, partial fractions, trigonometric integrals, trigonometric substitutions, improper integrals and i want to fully understand them. I've been working on problems from the book but can someone just give a simple explanation of how i know which type of integration to use for a certain problem (so i don't spend a lot of time trying to figure it out on the test) and maybe a few slightly complicating examples. or i can list an exercise from the book.

thanks

 

[Jameson]

Knowing what technique to use is something that comes with practice. The more you do, the more quickly you will see how to approach it.

For the trig integrals, be very familiar with the standard forms of sine, cosine and tangent inverse integrals. Here is a good page that lists a lot of general forms. If you see some terms that could be part of a right triangle: $\sqrt{a^2-x^2}$, $1+u^2$ or $a^2+u^2$ then that's a big hint that you should try a trig inverse function.

If you have to integrate a term with two strange things multiplied together and you notice that it's messy, but if you took the derivative of one of those pieces it would be less messy, then you could try integration by parts. For example:

$$\int xe^{6x} \,dx$$

This doesn't have an easy substitution and isn't a standard form for an integral right off the bat, but if we took the derivative of $x$, we could easily integrate $$e^{6x}$$, so integration by parts could work (and it does).

Hopefully this helps get you thinking. Let us know if you have more abstract or concrete questions.

 

[ineedhelpnow]

how would i integrate by parts if I am given something like $\int \ \frac{xe^{2x}}{(1+2x)^2},dx$

 

[MarkFL]

how would i integrate by parts if I am given something like $\int \ \frac{xe^{2x}}{(1+2x)^2},dx$

I wouldn't use integration by parts here:

$$I=\int\frac{xe^{2x}}{(2x+1)^2}\,dx=\int\frac{2(8x+4)e^{2x}-e^{2x}(8)}{\left(4(2x+1)\right)^2}\,dx$$

The integrand is now the derivative of a quotient...

 

[ineedhelpnow]

the only one we're allowed to use for this one is IBP

 

[ineedhelpnow]

all i need to know pretty much is how to integrate a fraction by parts

 

[MarkFL]

the only one we're allowed to use for this one is IBP

Try:

$$u=xe^{2x}\,\therefore\,du=(2x+1)e^{2x}\,dx$$

$$dv=\frac{1}{(2x+1)^2}\,dx\,\therefore\,v=-\frac{1}{2(2x+1)}$$

 

[ineedhelpnow]

im getting a wrong answer.

 

[Prove It]

im getting a wrong answer.

Then SHOW us what you have done, where you got stuck, and what you got as your answer. Then we might be able to give you some better guidance.

Mark is right, his choice of u and dv works.

 

[ineedhelpnow]

all i did was use the substitutions and i ended up getting $\frac{xe^{2x}}{4x+4}-\frac{e^{2x}}{4}$

 

[Prove It]

all i did was use the substitutions and i ended up getting $\frac{xe^{2x}}{4x+4}-\frac{e^{2x}}{4}$

Why are you using substitutions at all? Mark told you that Integration by Parts is the method to use, and even told you what to use as u, v, du and dv.

 

[ineedhelpnow]

I know what Mark told me ok? That's what I did and I did it just like he said but I'm doing something wrong.

@MarkFL i used $uv-\int \ vdu$ but I don't think my answer is right

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wait a sec. that answer i wrote was the one i got earlier before i asked. oops.

 

[Prove It]

I know what Mark told me ok? That's what I did and I did it just like he said but I'm doing something wrong.

@MarkFL i used $uv-\int \ vdu$ but I don't think my answer is right

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wait a sec. that answer i wrote was the one i got earlier before i asked. oops.

Well considering that in a thread called "techniques of integration" you say you used substitutions, which is a method of integration itself, instead of a different method (which you then say you used), then do you really wonder why you are being told to follow instructions?

Like I have said SEVERAL times, SHOW all your working out. THEN we will know where you have made your mistake!

 

[ineedhelpnow]

there's supposed to be a negative sign in front of $\frac{xe^{2x}}{2(2x+1)}$

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i don't get what you're talking about. i asked about the techniques of integration. then i asked how do i integrate fraction by parts because i didnt understand. he told me what to do and i did it. and I am getting it wrong for some reason ok? there arent really any steps. all i had to do was plug in the u v and du.

 

[MarkFL]

What Prove It is getting at, is that we can't really help you to find your error unless you show us what you did. If you just say I got this or I got the wrong answer, we can't know where you went wrong.

Post what you did, and we'll guide you from there. :D

 

[ineedhelpnow]

ok. he's right. sorry. ill post my steps. $-\frac{xe^{2x}}{2(2x+1)}-\int \ \frac{(2x+1)e^{2x}}{2(2x+1)},dx$
$=-\frac{xe^{2x}}{2(2x+1)}-\frac{1}{2}\int \ e^{2x},dx$
$=-\frac{xe^{2x}}{4x+4}-\frac{e^{2x}}{4}$

where did i go wrong?

 

[MarkFL]

ok. he's right. sorry. ill post my steps. $-\frac{xe^{2x}}{2(2x+1)}-\int \ \frac{(2x+1)e^{2x}}{2(2x+1)},dx$
$=-\frac{xe^{2x}}{2(2x+1)}-\frac{1}{2}\int \ e^{2x},dx$
$=-\frac{xe^{2x}}{4x+4}-\frac{e^{2x}}{4}$

where did i go wrong?

In your first step, the integral has the wrong sign...do you see why?

 

[ineedhelpnow]

:p oh now i see it. it's supposed to be + . is the sign the only that's wrong with it?

 

[Prove It]

:p oh now i see it. it's supposed to be + . is the sign the only that's wrong with it?

Also 2(2x+1) = 4x + 2, NOT 4x + 4...

 

[ineedhelpnow]

Also 2(2x+1) = 4x + 2, NOT 4x + 4...

double oops. that's embarrassing. besides that is the answer right?

 

[MarkFL]

double oops. that's embarrassing. besides that is the answer right?

Correct your errors, and then combine terms (don't forget the constant of integration) and see what you get. Take the derivative of your result, and if you get the original integrand then you know you have gotten the correct answer. :D

 

[ineedhelpnow]

its right! thanks Mark/Prove It!