Question

integration of cosdthita / 4-9sin²
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Answer 1

Answer:

Hello,

∫ [sinx cosx /(4cos²x + 9sin²x)] dx =

let's split 9sin²x into 4sin²x + 5sin²x:

∫ [sinx cosx /(4cos²x + 4sin²x + 5sin²x)] dx =

∫ {sinx cosx /[4(cos²x + sin²x) + 5sin²x]} dx =

(applying the identity cos²x + sin²x = 1)

∫ {sinx cosx /[4(1) + 5sin²x]} dx =

∫ [sinx cosx /(4 + 5sin²x)] dx =

let's divide and multiply by 10 obtaining in the numerator the derivative of the denominator:

(1/10) ∫ [10sinx cosx /(4 + 5sin²x)] dx =

(1/10) ∫ d(4 + 5sin²x) /(4 + 5sin²x) =

ending with:

(1/10) ln (4 + 5sin²x) + C