sin-4x-4sin-4-x-4sin-2-x-1-dx-

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

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

Answered by MJS_new last updated on 16/Mar/22

\frac{\sin 4 x}{1 - {4 \sin}^{2} x + {4 \sin}^{4} x} = \frac{2 \sin 4 x}{1 + \cos 4 x} = \frac{2 \sin 2 x \cos 2 x}{\cos^{2} 2 x} =

= 2 \tan 2 x

2 \int \tan 2 x d x = - \ln ∣ \cos 2 x ∣ + C

Answered by cortano1 last updated on 17/Mar/22

Another way

Z = \int \frac{\sin 4 x}{{({2 \sin}^{2} x - 1)}^{2}} dx

Z = \int \frac{\sin 4 x}{{(- \cos 2 x)}^{2}} dx

Z = \int \frac{\sin 4 x}{(\frac{1 + \cos 4 x}{2})} dx

Z = - \frac{1}{2} \int \frac{d (1 + \cos 4 x)}{1 + \cos 4 x}

Z = - \frac{1}{2} \ln ∣ 1 + \cos 4 x ∣ + c

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