Respuesta :

caylus
Hello,

[tex]\boxed{sin^4\ x= \frac{3}{8}- \frac{cos\ 2x}{2}+\frac{sin\ 4x}{8} }\\ Let's assume t=6x \ dt=6\ dx\\ \int\limits {sin^4\ 6x} \, dx = \frac{1}{6} *\int\limits {sin^4\ t} \, dt \\ = \dfrac{1}{6} *( \frac{3t}{8} - \frac{sin\ 2t}{4} +\frac{sin\ 4t}{32})+C\\ = \dfrac{3x}{8} - \frac{sin\ 12x}{24} +\frac{sin\ 24x}{192} +C [/tex]
The question is asking to calculate and find the integral of sin^4(6x)dx, and base on my further computation and formulation about the said equation, I would say that the integral value of the said equation is | | = 1/192 (72 x-8 sin(12 x)+sin(24 x))+constant. I hope you are satisfied with my answer and feel free to ask for more