How do you find the integral of ${cos}^{3} (2 x) \cdot sin (2 x) d x$ $cos^3(2x)*sin(2x)dx$ ?

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Answer 1 · 2016-09-23T20:15:22

$= - \frac{1}{8} {cos}^{4} 2 x + C$ $= - 1/8 cos^4 2x + C$

start by noticing that, from the power and chain rules:

$\frac{d}{d x} ({cos}^{4} 2 x) = 4 {cos}^{3} 2 x \cdot (- sin 2 x) \cdot 2$ $d/dx (cos^4 2x) = 4 cos^3 2x * ( - sin 2x ) * 2$

$= - 8 {cos}^{3} 2 x sin 2 x$ $= -8 cos^3 2x sin 2x$

and so your integral is actually

$\int \frac{d}{d x} (- \frac{1}{8} {cos}^{4} 2 x) d x$ $int d/dx ( - 1/8 cos^4 2x) dx$

$= - \frac{1}{8} \int \frac{d}{d x} ({cos}^{4} 2 x) d x$ $= - 1/8 int d/dx ( cos^4 2x) dx$

$= - \frac{1}{8} {cos}^{4} 2 x + C$ $= - 1/8 cos^4 2x + C$

How do you find the integral of cos3(2x)⋅sin(2x)dxcos^3(2x)*sin(2x)dx?