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# How to integrate?

$\int\frac{cos^{2}(x)+sin^{2}(x)}{1+sin^{4}(x)+cos^{4}(x)} dx$

Note by Majed Musleh
1 year, 4 months ago

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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. · 1 year, 4 months ago

Nice !

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

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. · 1 year, 4 months ago