OpenStudy (shamim):

integrate dx/(1-cos3x)

4 years ago
OpenStudy (anonymous):

using u-substitution u = 3x du = 3dx this will leave you with \[\frac{ 1 }{ 3 }\int\limits_{}^{}\frac{ 1 }{ 1-\cos(u)}du\]

4 years ago
OpenStudy (shamim):

hm

4 years ago
OpenStudy (shamim):

m i correct

4 years ago
OpenStudy (shamim):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

@shamim close but not quite

4 years ago
OpenStudy (shamim):

trying to understand

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (shamim):

i solved it in my way

4 years ago
OpenStudy (shamim):

result is as u

4 years ago
OpenStudy (shamim):

thanks for help me

4 years ago