Question

integrate dx/(1-cos3x)

Answer 1

using u-substitution

u = 3x
du = 3dx

this will leave you with

\[\frac{ 1 }{ 3 }\int\limits_{}^{}\frac{ 1 }{ 1-\cos(u)}du\]

Answer 2

hm

Answer 3

m i correct

Answer 4

\[-\frac{ 3 }{ 8 } x ^{2} \cot \frac{ 3x }{ 2 }\]

Answer 5

then substitute s = tan(u/2)
and \[ds = \frac{ 1 }{2 }du*\sec^{2}(\frac{ u }{ 2})\]

\[\sin(u) = \frac{ 2s }{ 1+s^{2} }\]

\[\cos(u) = \frac{ 1-s^{2} }{ 1+s^{2} }\]

\[du = \frac{ 2ds }{ 1+s^{2} }\]


leaving us with

\[\frac{1}{3}\int\limits_{}^{}\frac{ 2 }{(1+s^{2})(1-\frac{ 1-s^{2} }{ 1+s^{2} }) }ds\]


simply that to give you

\[\frac{ 1 }{ 3 }\int\limits\limits\limits_{}^{}\frac{ 1 }{ s^{2} }ds = -\frac{ 1 }{ 3s }+C\]

substitute in s = tan(u/2)

\[-\frac{ 1 }{ 3 }\cot(\frac{ u }{ 2 })+C\]

substitute in u = 3x

\[\frac{ 1 }{ 3 }\cot(\frac{ 3x }{ 2 })+C\]

Answer 6

@shamim close but not quite

Answer 7

trying to understand

Answer 8

think of it like using u-substitution twice... except i use 's' instead of 'u' again

Answer 9

i solved it in my way

Answer 10

result is as u

Answer 11

thanks for help me