With \(\cos^2 A + \sin^2 A = 1 \) and \(\sin(2A) = 2\sin A \cos A\), you can simplify the integrand to \( \frac2{4-\sin^2(2x)} \). Then multiply top and bottom by 2, substitute \(2\sin^2 (2x) = 1 - \cos(4x) \), then apply Tangent half angle substitution.

The numerator of the integrand can be factored into \( (\cos x + \sin x)(1 - \cos x \sin x) \). The denominator can be simplified like the previous one, then multiply top and bottom by 2, you're left with \( \frac{ ( \cos x + \sin x)(2 - 2\sin x \cos x)}{4 - \sin^2 (2x)} = \frac{ ( \cos x + \sin x)(2 - \sin(2x))}{(2 - \sin (2x))(2 + \sin(2x))} = \frac{\cos x + \sin x}{2 + \sin(2x)} = \frac{\cos x + \sin x}{3 - (\sin x - \cos x)^2} \). Set \(y = \sin x - \cos x \), then you're left with a familiar integral \( \int \frac{dz}{a^2-z^2} \), in this case \(a = \sqrt 3 \). Finish it off via partial fraction - cover up rule, and substitute everything back.

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TopNewestWith \(\cos^2 A + \sin^2 A = 1 \) and \(\sin(2A) = 2\sin A \cos A\), you can simplify the integrand to \( \frac2{4-\sin^2(2x)} \). Then multiply top and bottom by 2, substitute \(2\sin^2 (2x) = 1 - \cos(4x) \), then apply Tangent half angle substitution.

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Nice !

\[\int\frac{cos^{3}(x)+sin^{3}(x)}{1+sin^{4}(x)+cos^{4}(x)} dx\]

What about this ?

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The numerator of the integrand can be factored into \( (\cos x + \sin x)(1 - \cos x \sin x) \). The denominator can be simplified like the previous one, then multiply top and bottom by 2, you're left with \( \frac{ ( \cos x + \sin x)(2 - 2\sin x \cos x)}{4 - \sin^2 (2x)} = \frac{ ( \cos x + \sin x)(2 - \sin(2x))}{(2 - \sin (2x))(2 + \sin(2x))} = \frac{\cos x + \sin x}{2 + \sin(2x)} = \frac{\cos x + \sin x}{3 - (\sin x - \cos x)^2} \). Set \(y = \sin x - \cos x \), then you're left with a familiar integral \( \int \frac{dz}{a^2-z^2} \), in this case \(a = \sqrt 3 \). Finish it off via partial fraction - cover up rule, and substitute everything back.

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