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anonymous
 5 years ago
\[\int\limits_{}^{}(\cos x \cos 2x) dx/1cosx\]
anonymous
 5 years ago
\[\int\limits_{}^{}(\cos x \cos 2x) dx/1cosx\]

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is it \[\int\limits_{}^{}{\cos x\cos 2x \over 1\cos x}dx?\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the major problem is to simplify the trig function integration would be easy peasy.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, synthetic division to simplify the expression to: \[\cos x+{\cos^2 x\cos2x \over 1\cos x}\] But, cos^2xcos2x=sin^2x. So, it can simplified to: \\[[\cos x+{\sin^2x \over 1\cos x} =\cos x+{(1\cos x)(1+\cos x) \over 1 \cos x}=2\cos x+1\]x\] Easy peasy :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cos x  cos2x = cos x  (2cos^2 x 1) = cosx  cos^2x  cos^2x +1 = cos x(1cos x)  (1+cos x)(1cos x) = (1cos x)(cos x  1 cos x) = (1cos x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}(2\cos x+1)dx=2\sin x+x+c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is this the right answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea it is.. thanx man=)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait hold on.. the sign of sin should be negative

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I saw a problem you posted earlier that contains a^x, I couldn't solve. Are you sure about the problem?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea..if you can trust the book..lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i made some mistake in simplifying the numerator, but it simplifies to (1cos x)(1+2cos x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so your integral is : \[\int\limits_{}^{}(1+2cosx) dx\]
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