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Question Number 167438 by cortano1 last updated on 16/Mar/22

$$\int \frac{\sin 4\mathrm{x}}{4\sin {}^4\mathrm{x}-4\sin {}^2\mathrm{x}+1}\mathrm{dx}=?$$

Answered by MJS_new last updated on 16/Mar/22

$$\frac{\sin 4x}{1-{4\sin }^{2}x+{4\sin }^{4}x}=\frac{2\sin 4x}{1+\cos 4x}=\frac{2\sin 2x\cos 2x}{{\cos }^{2}2x}=$$$$=2\tan 2x$$$$2\int \tan 2xdx=-\ln \mid \cos 2x\mid +C$$

Answered by cortano1 last updated on 17/Mar/22

$$\mathrm{Another}\mathrm{way}$$$$\mathrm{Z}=\int \frac{\sin 4\mathrm{x}}{{({2\sin }^2\mathrm{x}-1)}^2}\mathrm{dx}$$$$\mathrm{Z}=\int \frac{\sin 4\mathrm{x}}{{(-\cos 2\mathrm{x})}^2}\mathrm{dx}$$$$\mathrm{Z}=\int \frac{\sin 4\mathrm{x}}{\left(\frac{1+\cos 4\mathrm{x}}{2}\right)}\mathrm{dx}$$$$\mathrm{Z}=-\frac{1}{2}\int \frac{\mathrm{d}(1+\cos 4\mathrm{x})}{1+\cos 4\mathrm{x}}$$$$\mathrm{Z}=-\frac{1}{2}\ln \mid 1+\cos 4\mathrm{x}\mid +\mathrm{c}$$$$$$

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