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I need to find the integral of sin(ax)cos(bx), please help

  • Thread starter crystal91
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  • #1
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Homework Statement


how to guess the value of the integral sinax cosbx and for what values of a and b is your formula valid?


Homework Equations



used the formula 1/2[(sin A-B)+sin(A+B)]

The Attempt at a Solution


this is what i got -cos(a+b)/2(a+b)-cos(a-b)/2(a-b).after that to guess the values of a and b i am lost.
Any help would be appreciated
thanx in advance!!
 

Answers and Replies

  • #2
VietDao29
Homework Helper
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Homework Statement


how to guess the value of the integral sinax cosbx and for what values of a and b is your formula valid?


Homework Equations



used the formula 1/2[(sin A-B)+sin(A+B)]

The Attempt at a Solution


this is what i got -cos(a+b)/2(a+b)-cos(a-b)/2(a-b).after that to guess the values of a and b i am lost.
Any help would be appreciated
thanx in advance!!
I'm not pretty sure if I understand your problem correctly. But what if a = b, or a = -b? Is your work still valid? :wink:
 
  • #3
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do u mean to say set b as a in the formula sinax cos bx,so it would be sinax cosax??
 
  • #4
VietDao29
Homework Helper
1,423
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do u mean to say set b as a in the formula sinax cos bx,so it would be sinax cosax??
Yup, if [tex]a \neq \pm b[/tex], then your work is correct.

You should also consider 2 special cases where [tex]a = \pm b[/tex].
 
  • #5
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then what do i do with this formula -cos(a+b)/2(a+b)-cos(a-b)/2(a-b)?
 
  • #6
VietDao29
Homework Helper
1,423
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then what do i do with this formula -cos(a+b)/2(a+b)-cos(a-b)/2(a-b)?
What do you mean by saying 'what do i do with this formula'? Your work is valid as long as [tex]a \neq \pm b[/tex]. When a is either b, or -b, the denominator of [tex]\frac{\cos[(a - b)x]}{a - b}[/tex] (and respectively, [tex]\frac{\cos[(a + b)x]}{a + b}[/tex]) becomes 0. So it's not valid anymore. You have to consider these 2 cases separately!!

By the way, you are forgetting x, and the constant of integration '+ C' in your result.
 
  • #7
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:confused:,so i would do it like this :1/2integral sin(2ax) which would be -1/2cos(2a)*1/2 so then the answer would be -1/4acos(2a)? and forget about this -cos(a+b)/2(a+b)-cos(a-b)/2(a-b)?
 
  • #8
VietDao29
Homework Helper
1,423
2


:confused:,so i would do it like this :1/2integral sin(2ax) which would be -1/2cos(2a)*1/2 so then the answer would be -1/4acos(2a)? and forget about this -cos(a+b)/2(a+b)-cos(a-b)/2(a-b)?
I suggest you look up the words: 'as long as', 'separately', and 'special cases' in the dictionary. Did you read my post above thoroughly enough??

Your work is valid IF [tex]a \neq \pm b[/tex].

When [tex]a = \pm b[/tex], it's no longer valid. And you should treat these as special cases. You have to divide into 3 cases:
  • Case 1: [tex]a \neq \pm b[/tex]
    Your work here...
  • Case 2: [tex]a = b[/tex]
    ...​
  • Case 3: [tex]a = -b[/tex]
    ...​

-----------------------

And by the way: [tex]\int \cos(ax) dx = \frac{1}{a} \sin (ax) + C[/tex] not [tex]\int \cos(ax) dx = \frac{1}{a} \sin a[/tex]
 

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