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

Try reducing the denominator to (sin x – cosx)^2 format.

Take sin x – cos x = t

Therefore sin x + cos x dx = dt

Then the question will be converted into dt/a^2 – t^2 format and hence we can apply the formula.

\(int frac{sinx+cosx}{3+sin2x} =int frac{sinx+cosx}{4-1+sin2x} \=int frac{sinx+cosx}{4-(1-sin2x)} =int frac{sinx+cosx}{4-(sin^{2}x+cos^{2}x-sin2x)} =int frac{sinx+cosx}{4-(sinx-cosx)^{2}} \let sin x- cos x =t => (cos x +sin x )dx = dt =int frac{dt}{4-t^2} = frac{1}{2*2} lnfrac{2+t}{2-t} = frac{1}{2*2} lnfrac{2+sinx-cosx}{2-sinx+cosx}\)


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