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

[crystal91]

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!

 

[VietDao29]

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?

 

[crystal91]

do u mean to say set b as a in the formula sinax cos bx,so it would be sinax cosax??

 

[VietDao29]

do u mean to say set b as a in the formula sinax cos bx,so it would be sinax cosax??

Yup, if a \neq \pm b, then your work is correct.

You should also consider 2 special cases where a = \pm b.

 

[crystal91]

then what do i do with this formula -cos(a+b)/2(a+b)-cos(a-b)/2(a-b)?

 

[VietDao29]

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 a \neq \pm b. When a is either b, or -b, the denominator of \frac{\cos[(a - b)x]}{a - b} (and respectively, \frac{\cos[(a + b)x]}{a + b}) 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.

 

[crystal91]

,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)?

 

[VietDao29]

,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 a \neq \pm b.

When a = \pm b, it's no longer valid. And you should treat these as special cases. You have to divide into 3 cases:

• Case 1: a \neq \pm b
  
  Your work here...​
• Case 2: a = b
  
  ...​
• Case 3: a = -b
  
  ...​

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

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