How do you find $\int {sin}^{2} (2 x) {cos}^{3} (2 x) d x$ $intsin^2 (2x) cos^3 (2x) dx$ ?

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Answer 1 · 2018-04-13T09:57:14

$\frac{{(sin 2 x)}^{3}}{6} - \frac{{(sin 2 x)}^{5}}{10} + C$ $(sin2x)^3/6-(sin2x)^5/10+C$

$\int {(sin 2 x)}^{2} \cdot {(cos 2 x)}^{3} \cdot d x$ $int (sin2x)^2*(cos2x)^3*dx$

= $\int {(sin 2 x)}^{2} \cdot {(cos 2 x)}^{2} \cdot cos 2 x \cdot d x$ $int (sin2x)^2*(cos2x)^2*cos2x*dx$

= $\int {(sin 2 x)}^{2} \cdot (1 - {(sin 2 x)}^{2}) \cdot cos 2 x \cdot d x$ $int (sin2x)^2*(1-(sin2x)^2)*cos2x*dx$

= $\frac{1}{2} \int {(sin 2 x)}^{2} \cdot (1 - {(sin 2 x)}^{2}) \cdot 2 cos 2 x \cdot d x$ $1/2int (sin2x)^2*(1-(sin2x)^2)*2cos2x*dx$

After using $y = sin 2 x$ $y=sin2x$ and $d y = 2 cos 2 y \cdot d y$ $dy=2cos2y*dy$ transforms, this integral became

$\frac{1}{2} \int y^{2} \cdot (1 - y^{2}) \cdot d y$ $1/2int y^2*(1-y^2)*dy$

= $\frac{1}{2} \int y^{2} \cdot d y - \frac{1}{2} \int y^{4} \cdot d y$ $1/2int y^2*dy-1/2int y^4*dy$

= $\frac{y^{3}}{6} - \frac{y^{5}}{10} + C$ $y^3/6-y^5/10+C$

= $\frac{{(sin 2 x)}^{3}}{6} - \frac{{(sin 2 x)}^{5}}{10} + C$ $(sin2x)^3/6-(sin2x)^5/10+C$

How do you find ∫sin2(2x)cos3(2x)dxintsin^2 (2x) cos^3 (2x) dx?