Is Integration by Parts Always the Best Method?

[rhey]

how do i integrate this??

a) ∫cot^(2)x cscx dx?

b) ∫[x^(3) + 2x^(2) + x]^(1/2) dx

 

[Suitengu]

I can tell you that the first problem requires you to use a trig identity followed by integration by parts and a little bit of manipulation.
The second one requires factoring, and then distributing and from there it should be quite easy.

Show me what you have done so far so we can take it from there.

 

[rhey]

first,
i've try to use
u=cotx n=2, but
du=-csc^(2)x dx

then, i use
u=csc x n=1, but
du=-csc x cotx dx

after that i don't know what to do! I'm stuck there!

 

[rocomath]

so step 1, integration by parts

let u=cotx, du=cscxcotxdx

now what do you get? and next step?

 

[Suitengu]

hmmm...substitution would not be ideal here as none is a derivative of the other and the derivative of cot(x) is -csc^2(x) not -cos^2(x).

What i rather you do is to use the trig identity that cot^2(x) = csc^2(x) - 1
and distribute to get csc^3(x) - csc(x). It will take more integration techniques to break this down further as I had said.

csc^3(x) needs to be integrated using integration by parts and then a simple rearrangement.

csc(x) should be done by multiplying it by (cscx + cotx)/(csx + cotx) and then using the general log rule to integrate.

Since i will have to leave for class, the second one;rocophysics' method is less tedious than the one i had pointed out so its best to go that way.

Just factor an x out first then factorize what you are left with. Then simplify and distribute then use the necessary integration rules to integrate.

later

 

[rhey]

Just factor an x out first then factorize what you are left with. Then simplify and distribute then use the necessary integration rules to integrate.

pls xhexk if i answered it correctly..

∫[x^(3) + 2x^(2) + x]^(1/2) dx

=∫{x[x^(2) + 2x +1]}^(1/2) dx

=∫ [x^(1/2)][(x+1)^(2)]^(1/2) dx

then the square root will be canceled out in [(x+1)^(2)]^(1/2)

=∫ [x^(1/2)](x+1) dx

=∫x^(1/2) * x dx + ∫ x^(1/2) dx

=∫ x^(3/2) dx + ∫ x^(1/2) dx

=(2/5) x^(5/2) + (2/3) x(3/2) + C

is that the final answer??

 

[Suitengu]

yup that's it right to the end mon. good job

 

[rhey]

the the first equation.. i don't te final answer but i have your idea..
let me try some..

∫cot^(2)x cscx dx

=∫ [csc^(2)x -1] cscx dx

=∫[csc^(3)x -cscx]dx

=∫csc^(3)x dx - ∫cscx dx

=∫ - ∫cscx dx *(cscx-cotx)/(cscx-cotx)

=∫ - ∫ [csc^(2)x -cscx cotx] dx / (cscx cotx)

=∫ - ln (cscx - cotx) +C

i don't know how to integrate cot^(3)x but I'm trying to figure it out

 

[Gib Z]

Perhaps an easier way to do a) is:

$$I = - \int \cot x ( - \cot x \csc x) dx$$.
Using The Pythagorean Identities to express cot x in terms of csc x, and letting u= csc x

$$I = \int \sqrt{u^2-1} du$$

That last integral is quite readily done by a hyperbolic trig substitution.

 

[rock.freak667]

b) ∫[x^(3) + 2x^(2) + x]^(1/2) dx

$$x^3+2x^2+x=x(x^2+2x+1)$$ factorise that quadratic again and...then put it back under the square root sign and see if anything occurs

 

[Suitengu]

ah rock freak I already told rhey to do that, and he/she got the answer. its the other one that's giving the problem now.

you would have to use integration by parts for csc^(3)x.

u = cscx dv = csc^(2)x dx
du = -cscxcotx dx v = -cotx

int(csc^(3)x)dx = -cscxcotx - int(cscx(csc^(2)x - 1))dx as cot^(2)x = csc^(2)x -1

2*int(csc^(3)x)dx = -cscxcotx + int(cscx)dx
int(csc^(3)x)dx = -(1/2)cscxcotx + (1/2)int(cscx)dx

and you can do int(cscx)dx as you did it earlier in the problem. that substitution method looks easier though. but this is the first thing that popped in mind.

 

[rock.freak667]

Integration by parts doesn't always give you an easier time than if you use a substitution...sometimes the easiest way is the best way