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anonymous
 5 years ago
How do I solve "integral [ sin(x) * cos(x) ] dx" ?
anonymous
 5 years ago
How do I solve "integral [ sin(x) * cos(x) ] dx" ?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{?}^{?} [ \sin(x) * \cos(x) ] dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Try usubstitution. Let u = sin(x). This means that du=cos(x)dx. Now it's ready to substitute back in the original equation to get: \[\int\limits_{?}^{?}udu\] This integral is (1/2)u^2 + c. Substituting back in for x you get: (1/2)sin^2(x) + c

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is it possible to do it with partial integration?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0also ... how do i know i have to do substitution? i mean ... i can easily get the antiderivative of both sin(x) and cos(x) ..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but getting an antiderivative of cos(x)sin(x) is different. You might be able to get it using parts but it would be more work than it's worth. When ever I do integrals I always ask myself first if I know an antiderivative and if I don't I move to usub because I find usub to be the next easiest.
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