∫(2cosx - 3sinx)/(6cosx+4sinx)
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Solution :
∴ ∫ {(2 cosx - 3 sinx)/(6 cosx + 4 sinx)} dx
= 1/2 ∫ {(2 cosx - 3 sinx)/(2 sinx + 3 cosx)} dx
= 1/2 ∫ {d(2 sinx + 3 cosx)}/(2 sinx + 3 cosx)
= 1/2 * log(2 sinx + 3 cosx) + C,
where C is the constant of integration
∴ the required integral is
= 1/2 * log(2 sinx + 3 cosx) + C.
Rule :
∫ [d{f(x)}]/f(x) = log{f(x)} + C,
where C is integral constant
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