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Integration Problem

  • Thread starter FallingMan
  • Start date
  • #1
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Homework Statement



Problem 1:
[PLAIN]http://img266.imageshack.us/img266/3852/problem1m.jpg [Broken]

Problem 2:
[PLAIN]http://img827.imageshack.us/img827/3862/problem3i.jpg [Broken]


2. The attempt at a solution

Prob. 1:

Well for this one I rewrote the integral first.

[tex]\int sec(x)(1-sin^2(x))dx[/tex]

Then I distributed the sec(x) term.

[tex]\int sec(x)-sec(x)sin^2(x)dx[/tex]

I'm stuck here :(

Prob. 2:

I think I'm supposed to use u-sub here.

Let [tex]u = \sqrt{x}[/tex]
[tex]du = \frac{1}{2\sqrt{x}}dx[/tex]
Then solve for [tex]dx = 2\sqrt{x}du[/tex]

Plug in for dx and you get:

[tex]\int (8x^2+2)(2\sqrt{x})(du)[/tex]

I'm really not sure if this is what's supposed to happen...

Thanks,
fm
 
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Answers and Replies

  • #2
eumyang
Homework Helper
1,347
10
Prob. 1:

Well for this one I rewrote the integral first.

∫sec(x) * (1-sin^2(x)) dx

Then I distributed the sec(x) term.

∫sec(x) - sec(x)*sin^2(x) dx

I'm stuck here :(
You're thinking too hard. Do you remember your trig identities? What does
[tex]1 - \sin^2 x[/tex] equal?

Prob. 2:

I think I'm supposed to use u-sub here.

Let u = sqrt(x), du = 1/2sqrt(x)dx
Then solve for dx = 2sqrt*du

Plug in for dx and you get:

∫(8x^2+2)*(2sqrt)*du

I'm really not sure if this is what's supposed to happen...

Thanks,
fm
Instead of doing that, write the fraction as a sum of two fractions:
[tex]\int \frac{8x^2 + 2}{\sqrt{x}} dx = \int \left( \frac{8x^2}{\sqrt{x}} + \frac{2}{\sqrt{x}} \right) dx = \int \frac{8x^2}{\sqrt{x}} dx + \int \frac{2}{\sqrt{x}} dx[/tex]
... and integrate the fractions separately.
 
  • #3
31
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You're thinking too hard. Do you remember your trig identities? What does
[tex]1 - \sin^2 x[/tex] equal?
Okay, I just looked it up. 1-sin^2(x) is equivalent to cos^2(x), right? That makes it easier then..

[tex] \int \frac{cos^2(x)}{cos(x)}dx = \int cos(x)dx = sin(x) + C[/tex]

:)?

Instead of doing that, write the fraction as a sum of two fractions:
[tex]\int \frac{8x^2 + 2}{\sqrt{x}} dx = \int \frac{8x^2}{\sqrt{x}}[ + \frac{2}{\sqrt{x}} dx[/tex]
... and integrate the fractions separately.
Okay so first part would be...

[tex] \frac{16}{5}x^\frac{5}{2} [/tex]

and second part would be...

[tex] 4\sqrt{x} [/tex]

So, adding them gives [tex]\frac{16}{5}x^\frac{5}{2}+4x^\frac{1}{2} + C[/tex]

Thank for for the help (I hope I did them right o_O)

By the way, how do you make the nice math symbols in your post? Any guide to how to do that?

EDIT: I think I got how to do it by viewing the edit options..
 
Last edited:
  • #4
eumyang
Homework Helper
1,347
10
#1 is right.

#2 is also right, but to nitpick, I personally don't like seeing an expression with fractional exponents and radicals mixed together. I would state the answer as
[tex]\frac{16}{5}x^{5/2} + 4x^{1/2} + C[/tex]
(and don't forget the C! :wink:)

EDIT: It's good that you're trying LaTex, but you got some errors. Check the code I used. As for a guide, you could look "ftp://ftp.ams.org/pub/tex/doc/amsmath/short-math-guide.pdf"[/URL].
 
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  • #5
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#1 is right.

#2 is also right, but to nitpick, I personally don't like seeing an expression with fractional exponents and radicals mixed together. I would state the answer as
[tex]\frac{16}{5}x^{5/2} + 4x^{1/2} + C[/tex]
(and don't forget the C! :wink:)

EDIT: It's good that you're trying LaTex, but you got some errors. Check the code I used. As for a guide, you could look "ftp://ftp.ams.org/pub/tex/doc/amsmath/short-math-guide.pdf"[/URL].[/QUOTE]

Thank for your help - I'm a real math noob so I might make another thread b/c I may have some more questions, lol.

I might also edit this post to try some equations and things out. :)
 
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