How do you find the integral of $(sin (5 x)) (5 cos (5 x)) 2$ $(sin(5x))(5cos(5x))2$ ?

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Answer 1 · 2015-01-31T18:23:59

The answer is: $- \frac{1}{2} cos 10 x + c$ $-1/2cos10x+c$ .

The double-angle formula says:

$sin (2 α) = 2 sin α cos α)$ $sin(2alpha)=2sinalphacosalpha)$ .

So: $2 sin 5 x cos 5 x = sin 10 x$ $2sin5xcos5x=sin10x$ .

Than the integral becomes:

$\int 5 sin 10 x d x = \frac{1}{2} \int 10 sin 10 x d x = - \frac{1}{2} cos 10 x + c$ $int5sin10xdx=1/2int10sin10xdx=-1/2cos10x+c$ ,

where I used the integration formula:

$intsinf(x)f'(x)dx=-cosf(x)+c$ .

How do you find the integral of (sin(5x))(5cos(5x))2(sin(5x))(5cos(5x))2(sin(5x))(5cos(5x))2?