how do you integrate ( 1 - cos x ) / [ cos x ( 1 +cos x) ]?

Answer 1

-sinx / sinx+sinx

Answer 2

separate the fractions>>>

step1>>1/[cos x(1+cos x)] - cos x/[cos x(1+cos x)]
step2>>(1/cos x) - (1/cos x + 1) - 1/(1 + cos x)
we know that 1/cos x = sec x
and 1 + cos x = 2cos^2(x/2)
put these in above to get the result
since integral of sec x = log|tan(x/2 + pi/4)|

Answer 3

Int[ (1-cosx)/ (cosx(1+cosx)) ]
We know that 1-cosx=2sin(x/2)^2 and 1+cosx=2cos(x/2)^2 and cosx=2sin(x/2)cos(x/2).(Half angle formula)
Plug it in,
=>Int[ {2sin(x/2)^2}.dx/ {2sin(x/2)cox(x/2)(2cos(x/2)^2)} ]
=>(1/2)Int[ tan(x/2).sec(x/2)^2 .dx ]----factor 2sin(x/2) is cancelled from numerator and denominator
=>Now substitute tan(x/2)=u,
=>(1/2).(sec(x/2)^2).dx=du
=>Back to original integral, Int[ u.du ]
=>(1/2)u^2 + C
=>(1/2)tan(x/2)^2 + C
Hint: Usually, when you have 1+cosx, 1-cosx in numerator or denominator or vice versa, its always easier to use half angle formula. Rationalisation gives higher powers of sinx, cosx, which are even more complex to solve. Unless you are confident solving all types of odd/even powers, its in your best interest to stick with half angles. Though working with half angles is confusing sometimes, it has advantage of having to work with less terms

Answer 4

seperate it to 1/cosx - 2/(1+cosx)

first part can be found in textbooks, it's calculated by partial integrating, and the other one is in form of 2/2Cos^2(x/2)=1+tan^2(x/2), its integral is easy=2tan(x/2)

Answer 5

by using integration

Answer 6

(a million - cos xe5056637c45f61f43a65a1e3d1ee2f71 + cos x) =[a million - {a million?2sin²(a million - {a million?2sin²(x/2)}]/[a million + {2cos²(x/2)?a million})}]/[a million + {2cos²(a million - {a million?2sin²(x/2)}]/[a million + {2cos²(x/2)?a million})?a million}] =2sin²(e5056637c45f61f43a65a1e3d1ee2f7e5056637c45f61f43a65a1e3d1ee2f7cos²(a million - {a million?2sin²(x/2)}]/[a million + {2cos²(x/2)?a million}) tan²(a million - {a million?2sin²(x/2)}]/[a million + {2cos²(x/2)?a million}) = sec²(a million - {a million?2sin²(x/2)}]/[a million + {2cos²(x/2)?a million})?a million subsequently ?(a million - cos xe5056637c45f61f43a65a1e3d1ee2f71 + cos x)dx=?{sec²(a million - {a million?2sin²(x/2)}]/[a million + {2cos²(x/2)?a million})?a million}dx =2tan(a million - {a million?2sin²(x/2)}]/[a million + {2cos²(x/2)?a million}) ? x+C

Answer 7

using trignometrical identities:
1-cosx=2sin^2(x/2) and 1+cosx=2cos^2(x/2) u get [tan^2(x/2)]/cosx
replacing cosx = {1-tan^2(x/2)}/[1+tan^2(x/2)]
u arrive at
tan^2(x/2).sec^2(x/2)dx/1-tan^2(x/2)
put tan(x/2)=u
taking differentials [sec^2(x/2)]/2.dx=du
integral[u^2]/{1-u^2}
ans is
1/2{1/2log[1+tan(x/2)] - tan(x/2)}
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[1-tan(x/2)]