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

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Answer 1 · 2015-01-31T18:23:59

The answer is: #-1/2cos10x+c#.

The double-angle formula says:

#sin(2alpha)=2sinalphacosalpha)# .

So: #2sin5xcos5x=sin10x#.

Than the integral becomes:

#int5sin10xdx=1/2int10sin10xdx=-1/2cos10x+c#,

where I used the integration formula:

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