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[TheDarkStar]

Well, integration by parts can be nested.

I have found that out already, yes. _{^{i hate it}}

When integrating by parts, how does one choose which part is used for the derivative and which part is used for the anti-derivative without trial and error? I keep picking the wrong sides for each problem and wind up having to erase everything and start a problem over, making homework take effectively twice as long. _{^{i hate it so much}}

If something looks hard/impossible to integrate, take its derivative instead and integrate the other part. If something looks like it would disappear if you repeatedly took its derivative, then take that expression's derivative. If both expressions will never disappear when integrating (such as with e^xcos(x)) you can still use integration by parts, but you have to do a bit more to get something useful (not spoiling it in case you don't know - it's better to try figuring it out first).

[frostshotgg]

For integration by parts, it's important to know which expressions will go to 0 fastest. Some expressions, the trig ones and exponentials will never go to 0 so you can just infinitely try to integrate by parts and get nowhere very quickly.

[Spehss _]

What can cos²(6x) be converted into using trig identities? For example, I know cos(2x)=cos²x-sin²x or cos(2x)=2cos²x-1 or etc.

No idea how to apply cos²(6x) instead of cos(2x.) It's a part of an integration problem (₀∫²cos²(6x)dx) and it's pretty much stumped me. _{^{One of many problems that I'm stuck on, actually.}}

[MagmaMcFry]

What can cos²(6x) be converted into using trig identities? For example, I know cos(2x)=cos²x-sin²x or cos(2x)=2cos²x-1 or etc.

No idea how to apply cos²(6x) instead of cos(2x.) It's a part of an integration problem (₀∫²cos²(6x)dx) and it's pretty much stumped me. _{^{One of many problems that I'm stuck on, actually.}}

Use the identity cos(2x)=2cos²x-1 with 6x instead of x and solve for cos²(6x), you'll get cos²(6x) = (cos(12x) + 1)/2.

[Spehss _]

Use a power reduction formula.

I have a reduction formula for the integral of sinⁿx but not cos.

Would the cos formula be ∫cosⁿxdx= -(1/n)sinxcos^n-1x+((n-1)/n)∫cos^n-2xdx ?

EDIT: Used that, it worked. Somehow. At this level of math I'm starting to think calculus is sufficiently advanced enough so that it is indistinguishable from magic.

[TheDarkStar]

Use a power reduction formula.

I have a reduction formula for the integral of sinⁿx but not cos.

Would the cos formula be ∫cosⁿxdx= -(1/n)sinxcos^n-1x+((n-1)/n)∫cos^n-2xdx ?

EDIT: Used that, it worked. Somehow. At this level of math I'm starting to think calculus is sufficiently advanced enough so that it is indistinguishable from magic.

I think Ispil meant this one: cos²u = (1 + cox 2u)/2. The formula you mentioned should work, too, but it will lead to a much larger number of integrals that need to be evaluated. It's a nice example of integration by parts, though.

[Spehss _]

And now for something completely different. Engineering physics question on vectors.

A vector is represented as 9.0i+6.0j-7.0k. What is the angle between the vector and the positive z axis?

I've calculated angles of 2d vectors (arctan(y/x)) but not 3d vectors. How do I solve this? I've tried arctan(z/x) which is arctan(-7/9) which results in -37.87^o. That's the angle between the x component and the z component, isn't it? How would I find the angle between the vector and the positive z axis? The answer in the back of the book says it's 123^o.

I think I'm misunderstanding something somewhere.

[MagmaMcFry]

Here's the insight you need: The dot product of any two euclidean vectors is equal to the product of their lengths multiplied by the cosine of the angle between them.

[Spehss _]

Here's the insight you need: The dot product of any two euclidean vectors is equal to the product of their lengths multiplied by the cosine of the angle between them.

So what I got out of that (I already knew this) was to treat the z axis as its own vector with a length of -7, calculate the length of the initial vector, which is ~12.9, then calculate arccos(-7/12.9) which results in 122.86^o or ~123^o.

[TheBiggerFish]

Tada!

That's your answer, isn't it?

[frostshotgg]

It's worth noting that while you put in a mathematically valid operation to get the right solution, where those numbers came from was wrong and it's pure coincidence that you got it right.

dotP(A,B) = Xa*Xb + Ya*Yb... = |A|*|B|*cos(theta)

In your problem, you would use A = (9,6,-7) and B = (0,0,1) (the easiest +Z vector to use). Then you'd find the dot product, which = 9*0 + 6*0 + -7*1 = -7. Then you'd find |A|*|B|, |A| = 12.9 like you said and |B| = 1, so |A|*|B| = 12.9, then you divide your dot product (-7) by the multiplied magnitudees (12.9) to find cos(theta), which is -.543, then you take acos(-.543) and get your 123.

[TheBiggerFish]

Yeah, I have no clue how to 3D vectors.

[Spehss _]

It's worth noting that while you put in a mathematically valid operation to get the right solution, where those numbers came from was wrong and it's pure coincidence that you got it right.

dotP(A,B) = Xa*Xb + Ya*Yb... = |A|*|B|*cos(theta)

In your problem, you would use A = (9,6,-7) and B = (0,0,1) (the easiest +Z vector to use). Then you'd find the dot product, which = 9*0 + 6*0 + -7*1 = -7. Then you'd find |A|*|B|, |A| = 12.9 like you said and |B| = 1, so |A|*|B| = 12.9, then you divide your dot product (-7) by the multiplied magnitudees (12.9) to find cos(theta), which is -.543, then you take acos(-.543) and get your 123.

I went and did some rereading of the textbook and some working with 2d vectors and see how the dot product is used for all that. Useful to know, thanks for pointing that out.

New question.
Two vectors:
d1=(4)i+(5)j-(6)k
d2=(-1)i+(2)j+(3)k

(a) What is the component of d1 along the direction of d2?
(b) What is the component of d1 that is perpendicular to the direction of d2 and within the plane of d1 and d2?

Spoiler: already solved a (click to show/hide)

I solved part (a) already using the dot component, so that d1*d2 = (-4)i+(10)j+(-18)k = -12

which means (magnitude of d1) * (magnitude of d2) * cos(angle between the two) = -12

so cos(angle)= -12/(d1*d2) or -12/(8.8*3.7) so angle = arccos(-12/(8.8*3.7)) = ~111

then 8.8cos(111)=~-3.2 which is the correct answer.

Don't know what to do with part B.

Tried adding 90 to the angle to make the direction perpendicular, which gives the component 8.8cos(201)=-8.2. The correct answer is positive 8.2, not negative 8.2. I think maybe the component is supposed to be the absolute value, and the negative just indicates it's in the opposite direction?

[jaked122]

I still don't get cross products. Is there any rule on how they scale up as you add dimensions to vectors?


I think that'd help a lot.

[frostshotgg]

Yeah. The dot product is super mega useful. The cross product really isn't. It's R3 only, serviceable in R2 by using 0 for Z.

Cross products are really weird, because they don't make much intuitive sense. You have an L shape, and the cross product of them is in the 3rd dimension. It always points "out" compared to the vectors it's a cross product of. You're finding the normal of the plane formed by those two vectors.