Integrate sin(x^(1/2)): Monday's Final Help Needed

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In summary, the conversation discusses solving an integral using substitution and integration by parts. The speaker recommends trying different techniques, such as guessing and playing around, and reading the textbook if needed. They then provide a step-by-step solution to the integral \int {\sin \left( {\sqrt x } \right)} dx using the substitution u = \sqrt x. The conversation concludes with the speaker expressing their gratitude for the help.
  • #1
bjon-07
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Can you please help me on this problem. I cannot seem to find the anwser.

Intergrate (sin(x^(1/2))

The problems tell me to do a subsition then use by parts to solve it.
 
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  • #2
Set [tex]u=\sqrt{x}[/tex]
 
  • #3
guessing and playing around is also a useful technique sometimes. just looking at that i would try xsin(x^(1/2)) just to see what comes out. well i got some stuff with x^(1/2) in front, which led me to try x^(1/2)cos(x^(1/2)). that came close enough to guess the rest.

guessing and playing around is actually easier and faster sometiems than keeping track of all the products and signs in the "parts" algorithm.
 
  • #4
When a question refers to certain style, i.e. "substitution", it is best to read the chapter to find out what it meant by it or atleast look it up in the textbook.

I recommend reading the sections of the chapters you have trouble with. If you don't read the text at all, may I ask why did you buy it?
 
  • #5
With questions which tell you to use a certain method I think it is best to give it a go even if you do not see where it might lead.

[tex]
\int {\sin \left( {\sqrt x } \right)} dx
[/tex]

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

From the substitution made earlier you can write: [tex]dx = 2udu[/tex]

So you now have:

[tex]
\int {\sin \left( {\sqrt x } \right)} dx
[/tex]

[tex]
= \int {\sin \left( u \right)2u} du
[/tex]

[tex]
= - 2u\cos \left( u \right) - \int {\left( { - \cos \left( u \right)} \right)} 2du
[/tex]

[tex]
= - 2u\cos \left( u \right) + \int {2\cos \left( u \right)} du
[/tex]

[tex]
= - 2u\cos \left( u \right) + 2\sin \left( u \right) + c
[/tex]

[tex]
= - 2\sqrt x \cos \left( {\sqrt x } \right) + 2\sin \left( {\sqrt x } \right) + c
[/tex]
 
  • #6
Thanks a lot for the help. Its all comming back to me now. I completey forgot that you can have a U in your du expression.
 

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