sin-4x-sin-4-x-cos-4-x-dx-

Question Number 128408 by bramlexs22 last updated on 07/Jan/21

ρ = \int \frac{\sin (4 x)}{\sin^{4} (x) + \cos^{4} (x)} d x

Answered by mr W last updated on 07/Jan/21

= \int \frac{\sin (4 x)}{1 - 2 \sin^{2} x \cos^{2} x} d x

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

= \int \frac{\sin (4 x)}{3 + \cos (4 x)} d (4 x)

= - \int \frac{1}{3 + \cos (4 x)} d (3 + \cos (4 x))

= - \ln [3 + \cos (4 x)] + C

Commented by bramlexs22 last updated on 07/Jan/21

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