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# Integration problems watch

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1. Lo, I'm having problems with these integration questions:

1. Substitution: integral of dx/(x^2)(1 + x^2)^1/2

If x = sinh u, dx = cosh u du

=> cosh u du/(sinh^2 u)(1 + sinh^2 u)^1/2

cosh^2 u - sinh^2 u = 1 => (1 + sinh^2 u)^1/2 = cosh u

=> cosh u du/cosh usinh^2 u

=> du/sinh^2 u

Have I done the substitution right? If so, how do I integrate cosech^2 u?

2. Parts: integral of (x^a)ln x dx (a is any real number)

= uv - integral of v(du/dx)

Taking u = ln x, dv/dx = (x^a) => du/dx = 1/x, v = (1/a + 1)x^(a + 1)

= (1/a + 1)ln x - (1/a + 1)integral of x^(a + 1).1/x dx

= (1/a + 1)ln x - [1/(a + 1)][(1/(a + 1))x^a + 1] + C

Evidently, I've done this wrong, cos it doesn't differentiate to what I started off with.

3. Parts again: ln [x + (1 + x^2)^1/2]

Took the ln term as being u, and dv/dx = 1 and I ended up with the integral of v(du/dx) being really nasty and not easy to integrate.

If anyone can help, thanx a lot.
2. (Original post by Nylex)
Lo, I'm having problems with these integration questions:

1. Substitution: integral of dx/(x^2)(1 + x^2)^1/2

If x = sinh u, dx = cosh u du

=> cosh u du/(sinh^2 u)(1 + sinh^2 u)^1/2

cosh^2 u - sinh^2 u = 1 => (1 + sinh^2 u)^1/2 = cosh u

=> cosh u du/cosh usinh^2 u

=> du/sinh^2 u

Have I done the substitution right? If so, how do I integrate cosech^2 u?

2. Parts: integral of (x^a)ln x dx (a is any real number)

= uv - integral of v(du/dx)

Taking u = ln x, dv/dx = (x^a) => du/dx = 1/x, v = (1/a + 1)x^(a + 1)

= (1/a + 1)ln x - (1/a + 1)integral of x^(a + 1).1/x dx

= (1/a + 1)ln x - [1/(a + 1)][(1/(a + 1))x^a + 1] + C

Evidently, I've done this wrong, cos it doesn't differentiate to what I started off with.

3. Parts again: ln [x + (1 + x^2)^1/2]

Took the ln term as being u, and dv/dx = 1 and I ended up with the integral of v(du/dx) being really nasty and not easy to integrate.

If anyone can help, thanx a lot.
yeeessss....no sorry i think only 'theone', elpaw or James F (in my year too!!) can help you here :-)
3. (Original post by Nylex)
= (1/a + 1)x^(a+1).ln x - (1/a + 1)integral of x^(a + 1).1/x dx
i don't know if that helps
4. (Original post by elpaw)
i don't know if that helps
Oops. I've got that in my written solution, just missed it out when typing, lol.

My answer for that one turns out to be:

(1/a + 1)[x^(a+1)]ln x - [1/(a + 1)^2]x^(a + 1) + C

It's definetly wrong .
5. (Original post by Nylex)
Lo, I'm having problems with these integration questions:

1. Substitution: integral of dx/(x^2)(1 + x^2)^1/2

If x = sinh u, dx = cosh u du

=> cosh u du/(sinh^2 u)(1 + sinh^2 u)^1/2

cosh^2 u - sinh^2 u = 1 => (1 + sinh^2 u)^1/2 = cosh^2 u

=> cosh u du/cosh^2 u sinh^2 u
=> du/(cosh u sinh²u)
6. (Original post by Nylex)
Lo, I'm having problems with these integration questions:

1. Substitution: integral of dx/(x^2)(1 + x^2)^1/2

If x = sinh u, dx = cosh u du

=> cosh u du/(sinh^2 u)(1 + sinh^2 u)^1/2

cosh^2 u - sinh^2 u = 1 => (1 + sinh^2 u)^1/2 = cosh u

=> cosh u du/cosh usinh^2 u

=> du/sinh^2 u

Have I done the substitution right? If so, how do I integrate cosech^2 u?

2. Parts: integral of (x^a)ln x dx (a is any real number)

= uv - integral of v(du/dx)

Taking u = ln x, dv/dx = (x^a) => du/dx = 1/x, v = (1/a + 1)x^(a + 1)

= (1/a + 1)ln x - (1/a + 1)integral of x^(a + 1).1/x dx

= (1/a + 1)ln x - [1/(a + 1)][(1/(a + 1))x^a + 1] + C

Evidently, I've done this wrong, cos it doesn't differentiate to what I started off with.

3. Parts again: ln [x + (1 + x^2)^1/2]

Took the ln term as being u, and dv/dx = 1 and I ended up with the integral of v(du/dx) being really nasty and not easy to integrate.

If anyone can help, thanx a lot.
tell me,wot do u get if u differentiate coth(x)...............
7. (Original post by IntegralAnomaly)
tell me,wot do u get if u differentiate coth(x)...............
coth(x)cosh(x)
8. (Original post by elpaw)
coth(x)cosh(x)
errr no u get -cosech^2(x), hmmm well if it isnt the integrand!
9. so the integral of cosech^2(u) is -coth(u)
10. (Original post by IntegralAnomaly)
errr no u get -cosech^2(x), hmmm well if it isnt the integrand!
sorry, was thinking of cosech (god all the h's are so confusing...)
11. (Original post by elpaw)
sorry, was thinking of cosech (god all the h's are so confusing...)
sooo true.....,whats worse is how u pronounce em.
12. (Original post by IntegralAnomaly)
sooo true.....,whats worse is how u pronounce em.
i al;ways pronounc them "sine haitch" "cos haitch" etc, and my teacher always had a go at me, saying they should be prononced "shine" and "cosh"
13. (Original post by elpaw)
i al;ways pronounc them "sine haitch" "cos haitch" etc, and my teacher always had a go at me, saying they should be prononced "shine" and "cosh"
u use to say sine-h-x? well it beats shine(x)
14. (Original post by IntegralAnomaly)
u use to say sine-h-x? well it beats shine(x)
yeah i did. whats wrong with that?
15. (Original post by elpaw)
yeah i did. whats wrong with that?
well for starters its not pronounced like that!
16. (Original post by IntegralAnomaly)
well for starters its not pronounced like that!
who says it isn't? who decided it shouold be 'shine'?
17. (Original post by elpaw)
who says it isn't? who decided it shouold be 'shine'?
the mathematician who invented it,anyway i dont blame u,ur a physicist.
in physics u dont care wot its name is just how it works!
18. (Original post by IntegralAnomaly)
so the integral of cosech^2(u) is -coth(u)
Thanx .

Note: Apparently ln [x + (1 + x^2)^1/2] = sinh^-1 x, so I'm assuming I can just do the integral of that by parts (ie. 1sinh^-1 x)

I've been told to pronounce the hyperbolic functions as shine and cosh as well, though I've heard people pronounce sinh x as sinesh x.
19. (Original post by Nylex)
Lo, I'm having problems with these integration questions:

2. Parts: integral of (x^a)ln x dx (a is any real number)

3. Parts again: ln [x + (1 + x^2)^1/2]

If anyone can help, thanx a lot.
2) integral of (x^a) ln(x) = (1/(a+1))ln(x)(x^(a+1)) - integral of x^a/(a+1). This should be okay. This gives the initial integral to be (lnx . x^(a+1))/(a+1) - x^(a+1)/(a+1)^2. I think this is what you got?

3) As you've said, just use parts on arsinh (x) .
20. (Original post by theone)
2) integral of (x^a) ln(x) = (1/(a+1))ln(x)(x^(a+1)) - integral of x^a/(a+1). This should be okay. This gives the initial integral to be (lnx . x^(a+1))/(a+1) - x^(a+1)/(a+1)^2. I think this is what you got?

3) As you've said, just use parts on arsinh (x) .
That is what I got. I couldn't differentiate last night and so thought it was wrong.

Thanx .

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