To find the equation: ∫2xcos⁡(3x)dx

Question

To find the equation: ∫2xcos⁡(3x)dx

Ayesha Gomez · Accepted Answer

We have to find the value of the integral
$\int 2 x \cos (3 x) d x$
Note that
$\int 2 x \cos (3 x) d x = 2 \cdot \int x \cos (3 x) d x$
Substitute $u = 3 x$ . Then we have $d u = 3 d x$
$= 2 \cdot \int (u \frac{\cos (u)}{9}) d u$ $[∵ x = \frac{u}{3} and d x = d \frac{u}{3}]$
$= 2 \cdot \frac{1}{9} \cdot \int u \cos (u) d u$
Recall: (Intergation By parts) $\int f (x) g (x) d x = f (x) \int g (x) d x - \int ((d \frac{f}{d x}) \int g (x) d x) d x$
[Apply Integration By Parts: $f (u) = u, g (u) = \cos (u)$

$= 2 \cdot \frac{1}{9} (u \sin (u) - \int \sin (u) d u)$

$= 2 \cdot \frac{1}{9} (u \sin (u) - (- \cos (u)))$

$= 2 \cdot \frac{1}{9} (3 x \sin (3 x) + (\cos (3 x)))$

$= \frac{2}{9} (3 x \sin (3 x) + \cos (3 x)) + C .$
Hence the final answer
$\int 2 x \cos (3 x) d x = \frac{2}{9} (3 x \sin (3 x) + \cos (3 x)) + C$

To find the equation: \int 2x\cos (3x)dx

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