What is the integral of $sin (2 x) {cos}^{2} (2 x) d x$ $sin(2x)cos^2(2x)dx$ ?

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Answer 1 · 2016-08-05T06:51:07

$\int sin (2 x) \cdot {cos}^{2} (2 x) d x = - \frac{1}{6} {cos}^{3} (2 x) + C$ $int sin(2x)*cos^2(2x) d x=-1/6cos^3(2x)+C$

$\int sin (2 x) \cdot {cos}^{2} (2 x) d x = ?$ $int sin(2x)*cos^2(2x) d x=?$

$Substitute u = cos (2 x); d u = - 2 sin (2 x) \cdot d x$ $"Substitute "u=cos(2x)" ; "d u=-2sin(2x)*d x$

$\int sin (2 x) \cdot {cos}^{2} (2 x) d x = - \frac{1}{2} \int u^{2} \cdot d u$ $int sin(2x)*cos^2(2x) d x=-1/2 int u^2*d u$

$So ; \int u^{n} d u = \frac{1}{n + 1} u^{n + 1}$ $"So ;" int u^n d u=1/(n+1)u^(n+1)$

$Apply n = 2$ $"Apply "n=2$

$\int sin (2 x) \cdot {cos}^{2} (2 x) d x = - \frac{1}{2} \cdot \frac{1}{3} u^{3}$ $int sin(2x)*cos^2(2x) d x=-1/2*1/3 u^3$

$Now ;undo substitution$ $"Now ;undo substitution"$

$\int sin (2 x) \cdot {cos}^{2} (2 x) d x = - \frac{1}{6} {cos}^{3} (2 x) + C$ $int sin(2x)*cos^2(2x) d x=-1/6cos^3(2x)+C$

What is the integral of sin(2x)cos2(2x)dxsin(2x)cos^2(2x)dx?