How do you integrate $\int 2 x sin 4 x d x$ $int 2x sin 4x dx$ ?

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Answer 1 · 2016-01-20T09:44:17

$\int 2 x sin (4 x) d x = \frac{1}{8} sin (4 x) - \frac{1}{2} x cos (4 x) + C$ $int2xsin(4x)dx = 1/8sin(4x)-1/2xcos(4x)+C$

Let $u = x$ $u = x$ and $d v = sin (4 x) d x$ $dv = sin(4x)dx$
Then $d u = d x$ $du = dx$ and $v = - \frac{1}{4} cos (4 x)$ $v = -1/4cos(4x)$

So, by the integration by parts formula $\int u d v = u v - \int v d u$ $intudv = uv - intvdu$

$2 \int x sin (4 x) d x = 2 (- \frac{1}{4} x cos (4 x) - \int ((- \frac{1}{4}) cos (4 x) d x)$ $2intxsin(4x)dx = 2(-1/4xcos(4x)-int((-1/4)cos(4x)dx)$

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

$= - \frac{1}{2} (x cos (4 x) - \frac{1}{4} sin (4 x) + C)$ $= -1/2(xcos(4x)-1/4sin(4x) + C)$

$= \frac{1}{8} sin (4 x) - \frac{1}{2} x cos (4 x) + C$ $= 1/8sin(4x)-1/2xcos(4x)+C$

How do you integrate ∫2xsin4xdxint 2x sin 4x dx?