Question

\[\int\limits_{}^{}(\cos x- \cos 2x) dx/1-cosx\]

Answer 1

Is it
\[\int\limits_{}^{}{\cos x-\cos 2x \over 1-\cos x}dx?\]

Answer 2

yes

Answer 3

the major problem is to simplify the trig function integration would be easy peasy.

Answer 4

Yeah, synthetic division to simplify the expression to:
\[\cos x+{\cos^2 x-\cos2x \over 1-\cos x}\]
But, cos^2x-cos2x=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 :)

Answer 5

cos x - cos2x = cos x - (2cos^2 x -1) = cosx - cos^2x - cos^2x +1
= cos x(1-cos x) - (1+cos x)(1-cos x) = (1-cos x)(cos x - 1 -cos x) = -(1-cos x)

Answer 6

\[\int\limits_{}^{}(2\cos x+1)dx=2\sin x+x+c\]

Answer 7

Is this the right answer?

Answer 8

yea it is.. thanx man=)

Answer 9

You're welcome!!

Answer 10

wait hold on.. the sign of sin should be negative

Answer 11

I saw a problem you posted earlier that contains a^x, I couldn't solve. Are you sure about the problem?

Answer 12

yea..if you can trust the book..lol

Answer 13

i made some mistake in simplifying the numerator, but it simplifies to (1-cos x)(1+2cos x)

Answer 14

so your integral is :
\[\int\limits_{}^{}(1+2cosx) dx\]