Indefinite Trignometric Integral

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Homework Help Overview

The discussion revolves around the indefinite integral ∫sinxcos(x/2)dx, which is being explored for test preparation rather than as a formal homework problem. Participants are examining trigonometric identities and substitution methods in the context of integral calculus.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss a substitution method involving u = ½ + ½ cosx and the resulting integral transformation. There is also mention of checking solutions graphically and concerns about discrepancies in results. Some participants suggest using product-to-sum identities as an alternative approach.

Discussion Status

The discussion is active, with participants providing feedback on each other's attempts and suggesting alternative methods. There is recognition of potential errors in the original approach, and participants are exploring different ways to express their mathematical work clearly.

Contextual Notes

Participants are working under the constraints of preparing for a test and are sharing insights on formatting mathematical expressions using LaTeX for clarity in their posts.

in the rye
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Homework Statement


∫sinxcos(x/2)dx

This isn't an actual homework problem, but one I found that I'm working on for test prep.

Homework Equations

The Attempt at a Solution


[/B]
∫sinxcos(x/2) dx = ∫sinx√((1+cosx)/2) dx

u = ½ + ½ cosx
-2 du = sinx dx

-2∫√(u) du = -2(2/3⋅u3/2) + c

-2(2/3⋅u3/2) + c = -4/3⋅[(1+cosx)/2)3/2] + c
-4/3⋅[(1+cosx)/2)3/2] = -4/3⋅(cos(x/2))3/2 + c

I tried to check my solution by graphing my answer with theirs, but the table is displaying different values.
 
Last edited:
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in the rye said:

Homework Statement


∫sinxcos(x/2)dx

This isn't an actual homework problem, but one I found that I'm working on for test prep.

Homework Equations

The Attempt at a Solution


[/B]
∫sinxcos(x/2) dx = ∫sinx√((1+cosx)/2) dx

u = ½ + ½ cosx
-2 du = sinx dx

-2∫√(u) du = -2(2/3⋅u3/2) + c

-2(2/3⋅u3/2) + c = -4/3⋅[(1+cosx)/2)3/2] + c
-4/3⋅[(1+cosx)/2)3/2] = -4/3⋅(cos(x/2))3/2 + c

I tried to check my solution by graphing my answer with theirs, but the table is displaying different values.
You have an error in the last line:
The formula you used for the substitution is ##\sqrt{\frac{1+\cos x}{2}}=\cos \frac{x}{2}##
 
Samy_A said:
You have an error in the last line:
The formula you used for the substitution is ##\sqrt{\frac{1+\cos x}{2}}=\cos \frac{x}{2}##

Gotcha, so it's probably easier to leave it where I had it in the second line. For some reason the solution isn't working, still. Even when I just plug the integral on my graphic calculator. :/

Is there a trick to getting my square roots to look fancy like yours so I can clean it up when I post?
 
there's another way, I think it is easier. Remember the product to sum trigonometric identities? Use it!
 
in the rye said:
Gotcha, so it's probably easier to leave it where I had it in the second line. For some reason the solution isn't working, still. Even when I just plug the integral on my graphic calculator. :/
Weird, I get the same result as you, except for the erroneous square root in the final result.

@MidgetDwarf 's suggestion is also good, but as said, you really are one symbol away from the solution.

in the rye said:
Is there a trick to getting my square roots to look fancy like yours so I can clean it up when I post?
Yes, I use ##\LaTeX##. It may look difficult if you never used it, but you'll learn it quickly. It makes posts with equations much more readable, and I highly recommend the use of it.

There is a guide here, there is also a link to that guide in the bottom left of the edit box.
If you see a complex ##\LaTeX## expression and want to see how it was composed, right-click the expression written in it and select "Show Math As -> TeX Commands". A small window will appear containing the code used for that expression. Or just quote a post: the ##\LaTeX## code will in general appear between ## or $$ tags.
 
Last edited:

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