# Need help with a couple of integrals

1. Feb 3, 2010

### noname1

Evaluate and simpfly each integral, specify u and du

∫ e^(x/3)

∫ 6 / sqrt(9x²-5)

∫ cosh(x) / sinh(x)

∫ 4x²cos(2x)

∫ (x³+4x) / (x²-4)

∫ (2^ln(x²)) / (x) from [1,2] and for this one i need to give a simplified symbolic(non decimal) answer

on the first one, its simple and its confusing cause cause of the (x/3), should it be the same thing?

Second i choose 9X²-5 as u, so du = 18x dx

so 6∫u^-2 making it 3x³-5x + C

Third, i choose u as sinh(x) making du = cosh(x) so the answer is log|sinh(x)|

Fourth is it possible to use the tabular method?

I got 4x²-2sin(2x)-8x-4cos(2x)+8sin(2x)+c

Fifth i can use the du/sqrt(u²-a²) = ln(u+sqrt(u²-a²) +C

ln(x+sqrtx²-2²) + C

the sixth one i am confused was wondering if someone could indicate what i should use for u and help me through it

And in addition verify if the others i did are correct

Thanks in advance

2. Feb 4, 2010

### tiny-tim

Hi noname1!
(try using the X2 tag just above the Reply box )
The question asks you to specify u and du.

If you mean is u = x/3, the answer is yes.
No, that u would be for ∫ 6x / sqrt(9x²-5) dx.

Your u needs to be a trig expression.
ok.
Try integration by parts.
uhh? the question says simplify the integral (if necessary, which it is here) … do that first.
Again, simplify it first.

3. Feb 4, 2010

### noname1

could you explain better on the simplifying part?

and on the first one i am getting (1/3)e^(x/3) but www.wolframalpha.com saying its 3e^(x/3)

i choose u as (x/3) making du/3 = dx

(1/3)∫ e^u = (1/3)e^(x/3)

aint i correct?

Last edited: Feb 4, 2010
4. Feb 4, 2010

### tiny-tim

(please use the X2 tag just above the Reply box )
No, u = x/3, du = dx/3.
(x³+4x) / (x²-4) is a fraction … simplify it.

2ln(x²) can also be simplified.

5. Feb 4, 2010

### noname1

i am having problems with the ∫ 4x²cos(2x)

u*v-∫ v*du

u = x² v = 1/2sin(2x)
du = 2x*dx dv = cos(2x)

4x²*1/2sin(2x) - 4∫1/2sin(2x)*2x dx

2x²sin(2x)-4∫sin(2x)*2x

u = x v = cos(2x)
du = dx dv = sin(2x)

2x²sin(2x)-4∫sin(2x)*2x +2xcos(2x) - 4∫cos(2x)

am i doing this correctly?

Last edited: Feb 4, 2010
6. Feb 4, 2010

### tiny-tim

ok so far

(but you would have found it a lot easier, and you'd be less likely to make mistakes, if you'd chosen u = 2x², v = 2cos(2x) …

then ∫ 4x²cos(2x) dx = [2x²*sin(2x)] - ∫ 4x*sin(2x) dx )

Sorry, this is just too confusing to read.

Try again, with brackets and using v = minus 2sin(2x)

7. Feb 4, 2010

### noname1

never mind i figured it out... by simplifying do you mean long division?

∫ (x³+4x) / (x²-4)

i dont remember much how you do but its something like this correct?

x²-4 | x³+4x
-x³+4x
+8x

Giving a result of x-8x? or should i factor out the x first from the equation

Last edited: Feb 4, 2010
8. Feb 4, 2010

### tiny-tim

You're less likely to make mistakes if you write it out like this …

(x³+4x)/(x²-4) = ((x³-4x)+8x)/(x²-4) = x + 8x/(x²-4)

9. Feb 4, 2010

### noname1

got that one too, now i just dont know how i can simplify the last one

10. Feb 4, 2010

### tiny-tim

Hint: what is eln(x²) ?

11. Feb 4, 2010

### noname1

hmmm, i am lost, i have no clue, can you give me the trig expression for the second one, i am just having problems on those 2.

And thanks for always trying to help me

12. Feb 5, 2010

### tiny-tim

(Are you serious?)

How is ln defined?

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