Problems with integration by parts

[Tjvelcro]

Homework Statement

Hi all! I seem to be having trouble doing integration by parts. I seem to have a pretty clear picture of the steps I need to do but something seems to always trick me. Usually I would ask my prof but she is away for a week.

I use the formula: uv - ∫vdu = given integration

Homework Equations

given integration = uv - ∫vdu

The Attempt at a Solution

Problem 1. Integrate! ∫(x^2) cosmx dx
u = x^2 -> du = 2x dx
dv = cosmxdt -> v = (1/m)sinmx

Use : uv - ∫vdu
(x^2) (1/m)sinmx - ∫(1/m)(sin(mx)) 2x dx
I pull the constants out
(x^2) (1/m)sinmx - (2 /m) ∫ (sin(mx)x) dx
I’m not sure what to do next… can I use the integration by parts again? Maybe using substitution?

Problem 2. Integrate! ∫t(sec(2t)^2) dt
U = t -> du = 1*dt
Dv = (sec(2t))^2 dt -> v = (½)(tan2t)
Use : uv - ∫vdu
t*(½)(tan2t) - ∫(½)(tan2t)dt
simplify a little
t/2(tan2t) – 1/2∫(tan2t)dt
I’m not sure what to do with the last term…. not sure how to integrate tan2t

Problem 3. Integrate! ∫(e^2x)sin3x dx
Not sure what to choose for u… usually I pick something that will simplify when I take the derivative. In this case (e^2x) does not simplify anything and sin3x just becomes 3cos3x which is more complex.

Any help would be appreciated!

Tjvelcro

 

[╔(σ_σ)╝]

Problem 1. Integrate! ∫(x^2) cosmx dx
u = x^2 -> du = 2x dx
dv = cosmxdt -> v = (1/m)sinmx

Use : uv - ∫vdu
(x^2) (1/m)sinmx - ∫(1/m)(sin(mx)) 2x dx
I pull the constants out
(x^2) (1/m)sinmx - (2 /m) ∫ (sin(mx)x) dx
I’m not sure what to do next… can I use the integration by parts again? Maybe using substitution?

Yeah use by parts for get rid of the x in ∫ (sin(mx)x)dx then simply integrate cos(mx).

Problem 2. Integrate! ∫t(sec(2t)^2) dt
U = t -> du = 1*dt
Dv = (sec(2t))^2 dt -> v = (½)(tan2t)
Use : uv - ∫vdu
t*(½)(tan2t) - ∫(½)(tan2t)dt
simplify a little
t/2(tan2t) – 1/2∫(tan2t)dt
I’m not sure what to do with the last term…. not sure how to integrate tan2t

∫(tan2t)dt is something you would find on an integral table, but I'm not a fan of them.

tanx = sinx / cos x

u = sinx
du = cosx dx

Remember cos^2(x) = 1- sin^2(x)

Problem 3. Integrate! ∫(e^2x)sin3x dx
Not sure what to choose for u… usually I pick something that will simplify when I take the derivative. In this case (e^2x) does not simplify anything and sin3x just becomes 3cos3x which is more complex.

This kind of integral usually just flips around and back since derievative of e^x and sinx still involve e^x and cosx and sinx respectively.

Set I = ∫(e^2x)sin3x dx

∫(e^2x)sin3x dx = (1/2)(e^2x)sin3x dx + 3/2 ∫(e^2x)cos3x dx

We can do the some thing with ∫(e^2x)cos3x dx and we will end up with our original integrand of ∫(e^2x)sin3x dx

 

[Char. Limit]

∫(tan2t)dt is something you would find on an integral table, but I'm not a fan of them.

tanx = sinx / cos x

u = sinx
du = cosx dx

That would be extremely difficult, considering you'd end with du in the denominator...

Try the substitution...

u=cos(x)
du = - sin(x) dx

instead. It will work out far better, and you won't need any more trig identities, except perhaps...

sec(x) = \frac{1}{cos(x)}

 

[╔(σ_σ)╝]

That would be extremely difficult, considering you'd end with du in the denominator...

Try the substitution...

u=cos(x)
du = - sin(x) dx

instead. It will work out far better, and you won't need any more trig identities, except perhaps...

sec(x) = \frac{1}{cos(x)}

What ?
∫(tan2t)dt

tanx = sinx / cos x

∫tanxdx

u = sinx
du = cosx dx
du/cosx =dx

∫u/(1-u^2) du -- This is a simple patial fractions.

 

[Char. Limit]

What ?
∫(tan2t)dt

tanx = sinx / cos x

∫tanxdx

u = sinx
du = cosx dx
du/cosx =dx

∫u/(1-u^2) du -- This is a simple patial fractions.

And my substitution gives a simple -1/u du. Integrate to -ln(u)+C and unsubstitute. No partial fractions at all.

 

[╔(σ_σ)╝]

Well great, you have an even quicker solution ! I was just a bit puzzled when you said my suggestion would be extremely difficult.

 

[Char. Limit]

Well great, you have an even quicker solution ! I was just a bit puzzled when you said my suggestion would be extremely difficult.

I neglected to consider the trig identity you used. TBH, I actually automatically go from \int tan(u) du to the solution.

 

[╔(σ_σ)╝]

Of course, everyone has a different way of approaching a problem. Heck, if you asked me on a different day ,this same question , I may provide a different substitution/solution .

I have to admit your solution is more enticing.