How do you find the integral of $({sin}^{2} (6 x)) ({cos}^{2} (6 x)) d x$ $(sin^2(6x))(cos^2(6x))dx$ ?

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How do you find the integral of $({sin}^{2} (6 x)) ({cos}^{2} (6 x)) d x$ $(sin^2(6x))(cos^2(6x))dx$ ?

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Answer 1 · 2016-06-04T18:38:47

You can rewrite this as follows

$\int ({sin}^{2} (6 x)) ({cos}^{2} (6 x)) d x = \int {(\frac{1}{2} \cdot 2 \cdot sin (6 x) \cdot cos (6 x))}^{2} d x = \int \frac{1}{4} {(sin 12 x)}^{2} d x$ $int(sin^2(6x))(cos^2(6x))dx=int (1/2*2*sin(6x)*cos(6x))^2dx= int 1/4(sin12x)^2 dx$

Remember that

$cos 2 x = {cos}^{2} x - {sin}^{2} x = 1 - 2 {sin}^{2} x \Rightarrow {sin}^{2} x = \frac{1}{2} \cdot (1 - cos 2 x)$ $cos2x=cos^2x-sin^2x=1-2sin^2x=> sin^2x=1/2*(1-cos2x)$

Hence $\frac{1}{4} \int ({sin}^{2} 12 x) d x = \int \frac{1}{8} (1 - cos 24 x) d x = \frac{x}{8} - \frac{1}{192} \cdot sin (24 x) + c$ $1/4int (sin^2 12x)dx=int 1/8(1-cos24x)dx=x/8-1/192*sin(24 x)+c$

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How do you find the integral of $({sin}^{2} (6 x)) ({cos}^{2} (6 x)) d x$ $(sin^2(6x))(cos^2(6x))dx$ ?

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How do you find the integral of $({sin}^{2} (6 x)) ({cos}^{2} (6 x)) d x$ $(sin^2(6x))(cos^2(6x))dx$ ?