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8 votes
8 votes

\int{[sin(9x)cos(9x)]} \, dx

User Nicholas Car
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1 Answer

22 votes
22 votes

Recall the double angle identity,

sin(2x) = 2 sin(x) cos(x)

Then we can write

sin(9x) cos(9x) = 1/2 sin(2 • 9x) = 1/2 sin(18x)

Then

∫ sin(9x) cos(9x) dx

= 1/2 ∫ sin(18x) dx

= -1/2 • 1/18 cos(18x) + C

= -1/36 cos(18x) + C

though you could continue with another double angle identity,

cos(2x) = cos²(x) - sin²(x)

to rewrite the antiderivative as

= -1/36 (cos²(9x) - sin²(9x)) + C

= 1/36 (sin²(9x) - cos²(9x)) + C

User Timur Catakli
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2.9k points