What are three different possible equivalent answers for the below? $\int$ $int$ sin(3x)cos(3x)dx

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$\int$ $int$ sin(3x)cos(3x)dx

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Answer 1 · 2018-02-21T05:31:29

$I = - \frac{1}{12} cos (6 x) + C$ $I = -1/12cos(6x) + C$

Here's how I would do it (1 approach). Letting $u = 3 x$ $u = 3x$ , we get $d u = 3 d x$ $du = 3dx$ and $d x = \frac{1}{3} d u$ $dx =1/3du$ .

$I = \frac{1}{3} \int sin (u) cos (u) d u$ $I = 1/3int sin(u)cos(u)du$

Now recall that $sin (2 α) = 2 sin α cos α$ $sin(2alpha) = 2sinalphacosalpha$ , thus

$I = \frac{1}{3} \int \frac{1}{2} sin (2 u) d u$ $I = 1/3int 1/2sin(2u) du$

$I = \frac{1}{6} \int sin (2 u) d u$ $I = 1/6int sin(2u) du$

Now we let $t = 2 u$ $t = 2u$ . Thus $d t = 2 d u$ $dt = 2du$ and $d u = \frac{d t}{2}$ $du = (dt)/2$ .

$I = \frac{1}{6} \int sin t \frac{d t}{2}$ $I = 1/6int sint (dt)/2$

$I = \frac{1}{12} \int sin t d t$ $I = 1/12 int sint dt$

This is a straightforward integration.

$I = - \frac{1}{12} cos t + C$ $I = -1/12cost + C$

Now just reverse your substitutions.

$I = - \frac{1}{12} cos (2 u) + C$ $I = -1/12cos(2u) + C$

$I = - \frac{1}{12} cos (6 x) + C$ $I = -1/12cos(6x) + C$

Hopefully this helps!