Find the integral of (sinx + cosx) / [3+sin2x] dx.

1.	Find the integral of (sinx + cosx) / [3+sin2x] dx.
Answer» The integral of (sinx + cosx) / [3+sin2x] dx is 1/4log {(2 + sinx - cosx) / (2 - sinx + cosx)} + C. Integration is exactly the reverse of differentiation. The integral of (sinx + cosx) / [3+sin2x] dx can be found using substitution method.

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