Integration by Parts (Solve ∫ x / cos2x dx) test on Monday please help!

[help meeeeeeee]

Hi right now im doing Integration by Parts and im stuck on a couple of examples, please help!

Solve ∫ x / cos²x dx

 

[ksdhart2]

Okay, so you say you're stuck on these examples. That implies that you've tried them and got stuck somewhere along the line. Please share with us all of your work on these problems, even if you know it's wrong. The more specific you can be about exactly what's giving you troubles, the better help we can provide you. Thank you.

 

[help meeeeeeee]

i was thinking about using u-substition at first but i dont think that will work so alright this is how i tried to solve it and i get nowhere by doing it:
∫ x / cos^2x dx.. i wrote it as ∫ 1/cos^2x*x dx
u= 1 / cos²x = (cosx)^-2
du = -2*(cosx)^-3 *2cosxsinx = -4sinx / (cosx)²
v = x²/2
dv= x

u*v - ∫v*du = x²/2cos²x - ∫-2x²sinx/(cosx)²
I have no idea what to do now

 

[ksdhart2]

i was thinking about using u-substition at first but i dont think that will work so alright this is how i tried to solve it and i get nowhere by doing it:
∫ x / cos^2x dx.. i wrote it as ∫ 1/cos^2x*x dx
u= 1 / cos²x = (cosx)^-2
du = -2*(cosx)^-3 *2cosxsinx = -4sinx / (cosx)²
v = x²/2
dv= x

u*v - ∫v*du = x²/2cos²x - ∫-2x²sinx/(cosx)²
I have no idea what to do now

What you've done is good so far, but you've made just a small error. You use u-substitution to set:

$\displaystyle \displaystyle u=\frac{1}{cos^2(x)}=(cos(x))^{-2}$

Then you try to use the chain rule to differentiate that. Well, let's make another substitution to illustrate what the chain rule's really doing. Let q = cos(x).

$\displaystyle \displaystyle \frac{du}{dx}=\frac{du}{dq} \cdot \frac{dq}{dx}$

$\displaystyle \displaystyle \frac{du}{dq}=-2q^3$

This is part where you went wrong. It appears to me as if you accidentally took the derivative of cos(2x) instead of cos(x).

$\displaystyle \displaystyle \frac{dq}{dx}=-sin(x)$

$\displaystyle \displaystyle \frac{du}{dx}=-2q^3 \cdot -sin(x)$

Re-substitute cos(x) for q:

$\displaystyle \displaystyle \frac{du}{dx}=-2cos^3(x) \cdot -sin(x)=\frac{2sin(x)}{cos^3(x)}=\frac{2tan(x)}{cos^2(x)}$

Now, using the integration by parts formula:

$\displaystyle \displaystyle \int \:udv= u \cdot v - \int \:vdu$

$\displaystyle \displaystyle \frac{1}{cos^2(x)} \cdot \frac{x^2}{2} - \int \:\frac{x^2}{2}\cdot \frac{2tan\left(x\right)}{cos^2\left(x\right)}dx$

Oh. Oops. Looks like that actually made the integral more complicated. If you wanted to continue, you could use integration by parts again, but it would get far worse before it got better. So, this suggests that the tactic you chose probably wasn't the best one. Unfortunately, the nature of integration by parts is such that you often try something and just end up spinning your wheels, unable to progress. That's a sign you should go back to the drawing board and try different substitutions. In this case, the important thing to remember is that reversing the "order" of the substitutions usually results in following a totally different process, and often one is much easier than the other. Try these substitutions and see where you get. It will go much more smoothly:

$\displaystyle u=x \implies du=?$

$\displaystyle dv=\dfrac{1}{cos^2(x)} \implies v=?$

 

[help meeeeeeee]

Hello ksdhart2 thank you for taking your time to help me^^

i changed it a bit so i put X as u because then du=1 so we dont have to repeat integration by parts. dv = (cosx)^-2 and therefore v = 1/sinxcosx.
SO now lets use the formula it gives me

x / sinxcosx - ∫ 1 / sinxcosx dx =
this is as far as i've gotten but this looks much more simple cuz i dont have to use part by integral again... BUT how do i find integral of 1/sinxcosx ( cuz thats equal to (sinxcosx)^-1 and to add 1 to its exponent makes it (sinxcosx)⁰ and that makes no sense..


EDIT: It says in my book that the answer is ln(cosx) + xtanx + C so perhaps looking at that someone could solve it and explain how to do it? Thanks!

 

[ksdhart2]

You're almost there. You definitely see what I mean about this method being far simpler to handle than before though, right? In any case, you seem to more or less understand the integration by parts, but it seems like a review of the basic trig derivatives and integrals might be in order, as you've messed up there again. I'm not actually 100% sure how you ended up with your answer for the integral of 1/cos^2(x), because that's actually a basic trig identity. Namely, we have this known fact:

$\displaystyle \displaystyle \frac{d}{dx}tan(x)=sec^2(x)=\frac{1}{cos^2(x)}$

Which means we have the inverse relationship:

$\displaystyle \displaystyle \int \:\frac{1}{cos^2\left(x\right)}dx=\int \:sec^2\left(x\right)dx=tan\left(x\right)+C$

If your textbook is anything like the one I used, I bet there's a table or list somewhere (mine was on the last few pages) of useful integrals and derivatives. The integral shown above can fairly easily be derived by using integration by parts, but it's such a well known integral that you'd be better off to just memorize it and skip doing the calculation in the future.

 

[help meeeeeeee]

Yeah thanks i just realized there was a list which said integral of 1/cos²x is tanx and when i derivate tanx i indeed get that integral but how do i find the integral without checking the list? because thats where i messed up in the first place right?D:

this is how i thought of finding the integral of 1/cos²x = (cosx)^-2 sooo add one to the exponent --> (cosx)^-1 the -1 falls down and also -sinx (derivative of cosx) and we dont wanna keep sinx so divide it by sinx --> (cosx)^-1/ sinx.

and (cosx)^-1 = 1/cosx sooo --> 1/ cosx/sinx = 1/cosxsinx...
what have i done wrong because obv this is not the same as tanx?

 

[ksdhart2]

Oh, okay. I see your line of thinking now. It certainly makes sense that there might be an equivalent to the product rule for integrals, but sadly it doesn't really work out. I've done some looking and it turns out I was mistaken before when I said you can derive the integral of sec²(x) by using integration by parts. I believe it's possible to do it that way, but it turns out it's not easy at all. If you're interested, there's a Stack Exchange thread discussing ways of proving the integral without already knowing its solution. I found it to be an interesting read, but taking the time to understand it might not be in your best interest right at this particular moment - probably best to study for your test instead I can say with about 99% certainty that you won't need to know this during the test.

 

[pka]

Hi right now im doing Integration by Parts and im stuck on a couple of examples, please help!
Solve ∫ x / cos²x dx

$\displaystyle \begin{cases}u=x & dv=(\cos^2(x))^{-1}dx=\sec^2(x)dx \\ du=dx & v=\tan(x)\end{cases}$
Thus
$\displaystyle uv - \int {\tan (x)dx = x\tan (x) + \log (\cos (x)) + c}$

 

[help meeeeeeee]

Thank you for helping me ^^ I appreciate it a lot, that example was getting in my head all day!
i dont think i'll need it for my test either but i might come back and read some later on ;D Thankss!