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Question Number 163008 by tounghoungko last updated on 03/Jan/22

$$Calculate$$$$\int \frac{(2-4\sin x\cos x)(1+\sin 2x)}{\sin {}^{4}2x+64\cos {}^{4}2x}dx$$$$$$

Answered by som(math1967) last updated on 03/Jan/22

$$\int \frac{2(1-2sinxcosx)(1+sin2x)}{{sin}^{4}x+64{cos}^{4}x}dx$$$$\int \frac{2(1-{sin}^{2}2x)dx}{{sin}^{4}x+64{cos}^{4}x}$$$$\int \frac{2{cos}^{2}2{xsec}^{4}2xdx}{{sec}^{4}2x({sin}^{4}2x+64{cos}^{4}2x)}$$$$\int \frac{2{sec}^{2}2xdx}{{tan}^{4}2x+64}$$$$lettan2x=z$$$$\therefore 2{sec}^{2}2xdx=dz$$$$\int \frac{dz}{z^4+8^2}$$$$\frac{1}{16}\int \frac{(8+z^2)+(8-z^2)dz}{z^4+8^2}$$$$\frac{1}{16}\int \frac{\frac{8+z^2}{z^2}}{\frac{z^4+8^2}{z^2}}dz-\frac{1}{8}\int \frac{\frac{z^2-1}{z^2}}{\frac{z^4+8^2}{z^2}}dz$$$$\frac{1}{16}\int \frac{d(z-\frac{8}{z})}{{(z-\frac{8}{z})}^2+{\left(4\right)}^2}-\frac{1}{16}\int \frac{d(z+\frac{8}{z})}{{(z+\frac{8}{z})}^2-{\left(4\right)}^2}$$$$\frac{1}{64}{\tan }^{-1}\left(\frac{z-\frac{1}{z}}{4}\right)-\frac{1}{128}ln\frac{z+\frac{1}{z}-4}{z+\frac{1}{z}+4}+C$$$$z=tan2x$$

Answered by Ar Brandon last updated on 03/Jan/22

$$I=\int \frac{(2-4\sin x\cos x)(1+\sin 2x)}{{\sin }^{4}2x+{64\cos }^{4}2x}dx$$$$=\int \frac{2(1-\sin 2x)(1+\sin 2x)}{{\sin }^{4}2x+{64\cos }^{4}2x}dx$$$$=2\int \frac{1-{\sin }^{2}2x}{{\sin }^{4}2x+{64\cos }^{4}2x}dx=2\int \frac{{\sec }^{2}2x}{{\tan }^{4}2x+64}dx$$$$t=\tan 2x,dt={2\sec }^{2}2xdx$$$$=\int \frac{dt}{t^4+64}=\frac{1}{16}\int \frac{(t^2+8)-(t^2-8)}{t^4+64}dt$$$$=\frac{1}{16}\int \frac{1+\frac{8}{t^2}}{{(t-\frac{8}{t})}^2+16}dt-\frac{1}{16}\int \frac{1-\frac{8}{t^2}}{{(t+\frac{8}{t})}^2-16}dt$$$$=\frac{1}{16}\int \frac{du}{u^2+16}-\frac{1}{16}\int \frac{d\mathrm{v}}{{\mathrm{v}}^2-16}$$$$=\frac{1}{64}{\tan }^{-1}\left(\frac{u}{4}\right)+\frac{1}{64}{\tanh }^{-1}\left(\frac{\mathrm{v}}{4}\right)+C$$$$=\frac{1}{64}{\tan }^{-1}\left(\frac{{\tan }^{2}2x-8}{4\tan 2x}\right)+\frac{1}{128}\ln (\mid \frac{{\tan }^{2}2x+4\tan 2x+8}{{\tan }^{2}2x-4\tan 2x+8}\mid )+C$$

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