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Originally posted by ND
After doing part i), i'm sure it'll become pretty obvious.
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Originally posted by ND
After doing part i), i'm sure it'll become pretty obvious.
let u=2sinx+cosxOriginally posted by ...
i got up to that
but what do i do after!
like whats u!
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boy i made a big fool of myself..thx...Originally posted by ND
No substitution is necessary... It's in the form f'(x)/f(x).
ok, say in another scenario..how would you make the u=??? or its not required in 3u?Originally posted by freaking_out
let u=2sinx+cosx
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.Originally posted by ...
...ok, say in another scenario..how would you make the u=??? or its not required in 3u?
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).Originally posted by ...
but how did u come up with u=2sinx+cosx ...not u=2cosx+sinx ??
There is no method of determining the substitution, you just do it by examination.Originally posted by ...
ok, say in another scenario..how would you make the u=??? or its not required in 3u?