# Math Help - integration

1. ## integration

hii!

how do you evaluate this integral.

thanks guys

2. Let w = sqr(x)

x = w^2

dx = 2wdw

integral [ (1/w) * 2w * sinw dw ]

the 2 w's cancel so your left with

integral [ 2 sin(w) dw ]

I think you can get the rest from there

3. (this one should be incorrect, ill check my notes later)

OR possibly

I think 2nd one is correct.

4. OR possibly

I think 2nd one is correct.

R_math... try using the substitute method.

5. I dunno why but I prefer the way I did it, provided I get it right of course.

6. Originally Posted by r_maths
I dunno why but I prefer the way I did it, provided I get it right of course.

I'm sorry to say but that wasn't right.

sine and cosine are functions, not constants. When you have sin(theta) that is not the same as sine * theta. If you have sin(theta)/theta, you cannot "reduce" the theta's and declare that sin(theta)/theta = sin

That doesn't make any sense.

In the future, I would recommend you do it the way smooth did, using substitution.

7. Originally Posted by smoothi963
Let w = sqr(x)

x = w^2

dx = 2wdw

integral [ (1/w) * 2w * sinw dw ]

the 2 w's cancel so your left with

integral [ 2 sin(w) dw ]

I think you can get the rest from there
Thank you I like that method.

= -2cosw ???

8. Thank you I like that method.

= -2cosw ???

-2cosw + c.....but remember you substituted the w, so you will need to rewrite wat w was.

-2cos(sqr(x)) + c

9. Originally Posted by ecMathGeek
I'm sorry to say but that wasn't right.

sine and cosine are functions, not constants. When you have sin(theta) that is not the same as sine * theta. If you have sin(theta)/theta, you cannot "reduce" the theta's and declare that sin(theta)/theta = sin

That doesn't make any sense.

In the future, I would recommend you do it the way smooth did, using substitution.
Thanks, i'll remember that next time.

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Enita, no need to thank me. I gave you wrong solution.