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

Homework Helper

## 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

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??

Homework Helper

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)?

Homework Helper

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)?

Homework Helper

,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When $$a = \pm b$$, it's no longer valid.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: $$a \neq \pm b$$
• Case 2: $$a = b$$
• Case 3: $$a = -b$$
And by the way: $$\int \cos(ax) dx = \frac{1}{a} \sin (ax) + C$$ not $$\int \cos(ax) dx = \frac{1}{a} \sin a$$