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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 !!! :)

  • maths -

    use what the first part showed. Using c for cos and s for sin, you showed that

    s+8c = 2(2s+c) + 3(2c-s)

    so,

    (s+8c)/(2s+c)
    = [2(2s+c) + 3(2c-s)]/(2s+c)
    = 2 + 3(2c-s)/(2s+c)

    now, you have ∫2 dx + ∫3du/u
    integral is

    2x + 3lnu
    = 2x + 3ln(2s+c) + C

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