How do you integrate $\int x cos 3 x d x$ $int x cos 3 x dx$ using integration by parts?

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How do you integrate $\int x cos 3 x d x$ $int x cos 3 x dx$ using integration by parts?

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Answer 1 · 2016-01-10T09:52:17

Step by step working is shown below on how to go about integration by parts.

Explanation:

$\int x cos (3 x) d x$ $intxcos(3x)dx$

Integration by parts rule is

$\int u d v = u v - \int v d u$ $int udv = uv-int vdu$

First we select $u$ $u$
Using a mnemonic ILATE which stands for Inverse, Logarithmic, Algebraic, Trigonometric and Exponential we select $u$ $u$ in the preferred order given by ILATE.

In our problem, we can select $u = x$ $u=x$ this is because Algebraic function is preferred above the Trigonometric.

$u = x$ $u=x$
Differentiating with respect to $x$ $x$
$d u = d x$ $du = dx$

Once $u$ $u$ is selected remaining terms for the $d v$ $dv$

$d v = cos (3 x) d x$ $dv=cos(3x)dx$

To find $v$ $v$ we have to integrate this function.

$v = \int d v = \int cos (3 x) d x$ $v=int dv = int cos(3x)dx$

$v = \frac{sin (3 x)}{3}$ $v=sin(3x)/3$

Using the integration part rule

$\int u d v = u v - \int v d u$ $int udv = uv-int vdu$

$\int x cos (3 x) d x = \frac{1}{3} x sin (3 x) - \int (\frac{sin (3 x)}{3}) d x$ $intxcos(3x)dx = 1/3xsin(3x) - int(sin(3x)/3)dx$

$\int x cos (3 x) d x = \frac{1}{3} x sin (3 x) + \frac{1}{9} cos (3 x) + C$ $intxcos(3x)dx = 1/3xsin(3x) +1/9cos(3x)+C$

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How do you integrate $\int x cos 3 x d x$ $int x cos 3 x dx$ using integration by parts?

1 Answer

Explanation:

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