Question

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Answer 1

$\int \dfrac{sinx\cdot cosx}{sin^4x+cos^4x}\, dx=\\\\\\\star \; \; sin^4x+cos^4x=(sin^4x+2\cdot sin^2x\cdot cos^2x+cos^4x)-2\cdot sin^2x\cdot cos^2x=\\\\=(\underbrace {sin^2x+cos^2x}_{1})^2-2\cdot sin^2x\cdot cos^2x=1-2sin^2x\cdot cos^2x=\\\\=1-2\cdot (sinx\cdot cosx)^2=1-2\cdot (\frac{1}{2}sin2x)^2=1-\frac{1}{2}\, sin^22x\; \; \star \\\\\\=\int\dfrac{\frac{1}{2}\, sin2x}{1-\frac{1}{2}\, sin^22x}\, dx=\dfrac{1}{2}\int \dfrac{sin2x\, dx}{\frac{2-sin^22x}{2}} =\int \dfrac{sin2x\, dx}{2-sin^22x}=$

$=\int \dfrac{sin2x\, dx}{2-(1-cos^22x)}=-\dfrac{1}{2}\int \dfrac{-2\, sin2x\, dx}{1+cos^22x}=-\dfrac{1}{2}\int \dfrac{d(cos2x)}{1+cos^22x}=\\\\\\=\Big[\; t=cos2x\; ,\; dt=-2sin2x\, dx\; ,\; \int \dfrac{dt}{1+t^2}=arctgt+C\; \Big]=\\\\\\=-\frac{1}{2}\cdot arctg(cos2x)+C\; .$