simple 3u..plz help (1 Viewer)

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Originally posted by ND
After doing part i), i'm sure it'll become pretty obvious.

 

[ND]

Ok, expand out then equate and you get A=2, B=3.
so:
2(2sinx+cosx)+3(2cosx-sinx)=sinx+8cosx

Now (sinx+8cosx)/(2sinx+cosx) = (2(2sinx+cosx)+3(2cosx-sinx))/(2sinx+cosx)
= 2 + (3(2cosx-sinx)/(2sinx+cosx))

Now do you know how to integrate that?

 

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i got up to that

but what do i do after!

like whats u!

 

[freaking_out]

Originally posted by ...
i got up to that

but what do i do after!

like whats u!

let u=2sinx+cosx

 

[ND]

No substitution is necessary... It's in the form f'(x)/f(x).

 

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holy lord...

Originally posted by ND
No substitution is necessary... It's in the form f'(x)/f(x).

boy i made a big fool of myself..thx...

Originally posted by freaking_out
let u=2sinx+cosx

ok, say in another scenario..how would you make the u=??? or its not required in 3u?

 

[freaking_out]

Originally posted by ...
...ok, say in another scenario..how would you make the u=??? or its not required in 3u?

hey, u wanted me to give a substitution (since its the 3u method), so thats what i gave u....otherwise i would've done it like NDs way.

 

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but how did u come up with u=2sinx+cosx ...not u=2cosx+sinx ??

 

[freaking_out]

Originally posted by ...
but how did u come up with u=2sinx+cosx ...not u=2cosx+sinx ??

well, i just saw the derivative of the denominator in the numerator, i.e f'(x)/f(x), and since this is the standard for log, then u just let u=(to denominator).

basically, its more or less trial and error type of thing, when u wanna find the substitution.

 

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oh darn...


oh well..thx a lot fellas..

ur help was greatly appreciated...

 

[ND]

Originally posted by ...

ok, say in another scenario..how would you make the u=??? or its not required in 3u?

There is no method of determining the substitution, you just do it by examination.