maths
posted by Candice
by equating the coefficients of sin x and cos x , or otherwise, find constants A and B satisfying the identity.
A(2sinx + cosx) + B(2cosx  sinx) = sinx + 8cosx
I got A = 2, B = 3, which the answers said were correct.
However, another part of the question asked
hence integrate
sinx + 8cosx
dx
2sinx + cosx
and the answers were 2x + 3Ln (2sinx + cosx) + C
Could you explain how to get this?
Thanks in advance !!! :)

Steve
use what the first part showed. Using c for cos and s for sin, you showed that
s+8c = 2(2s+c) + 3(2cs)
so,
(s+8c)/(2s+c)
= [2(2s+c) + 3(2cs)]/(2s+c)
= 2 + 3(2cs)/(2s+c)
now, you have ∫2 dx + ∫3du/u
integral is
2x + 3lnu
= 2x + 3ln(2s+c) + C
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