How do you integrate $\int x cos (2 x)$ $int xcos(2x)$ by integration by parts method?

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Answer 1 · 2016-08-19T13:59:49

$\frac{1}{4} (2 x \cdot sin 2 x + cos 2 x) + C$ $1/4(2x*sin2x+cos2x)+C$ .

Let $I = \int x cos 2 x d x$ $I=intxcos2xdx$

We use the Rule of Integration by Parts, which is,

$\int u v d x = u \int v d x - \int [\frac{d u}{d x} \int v d x] d x$ $intuvdx=uintvdx-int[(du)/dxintvdx]dx$

We take,

$u = x, s o \frac{d u}{d x} = 1; & v = cos 2 x, s o, \int v d x = \frac{sin 2 x}{2}$ $u=x, so (du)/dx=1; & v=cos2x, so, intvdx=(sin2x)/2$ .

Hence, $I = \frac{x}{2} \cdot sin 2 x - \frac{1}{2} \int sin 2 x d x$ $I=x/2*sin2x-1/2intsin2xdx$

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

$= \frac{1}{4} (2 x \cdot sin 2 x + cos 2 x) + C$ $=1/4(2x*sin2x+cos2x)+C$ .

Enjoy Maths.!

How do you integrate ∫xcos(2x)int xcos(2x) by integration by parts method?