∫cos(2x)sec(3x)

 
方法1.方法2. Mathematica
WolframAlpha["Integrate[Cos[2x]Sec[3x],x]", "PodCells", PodStates -> {"IndefiniteIntegral__Step-by-step solution"}][[2]]
Take the integral: cos ( 2 x ) sec ( 3 x ) d x Write   cos ( 2 x ) sec ( 3 x )   as   cos 2 ( x ) cos 3 ( x ) 3 sin 2 ( x ) cos ( x ) sin 2 ( x ) cos 3 ( x ) 3 sin 2 ( x ) cos ( x )   =   ( cos 2 ( x ) cos 3 ( x ) 3 sin 2 ( x ) cos ( x ) sin 2 ( x ) cos 3 ( x ) 3 sin 2 ( x ) cos ( x ) ) d x Integrate   the   sum   term   by   term   and   factor   out   constants:    =   sin 2 ( x ) cos 3 ( x ) 3 sin 2 ( x ) cos ( x ) d x + cos 2 ( x ) cos 3 ( x ) 3 sin 2 ( x ) cos ( x ) d x ⁠  Multiply   numerator   and   denominator   of   sin 2 ( x ) cos 3 ( x ) 3 sin 2 ( x ) cos ( x )   by   cot ( x ) csc 3 ( x )   =   cot ( x ) csc ( x ) 3 cot 2 ( x ) cot 4 ( x ) d x + cos 2 ( x ) cos 3 ( x ) 3 sin 2 ( x ) cos ( x ) d x ⁠  Prepare   to   substitute   u = csc ( x )   Rewrite   cot ( x ) csc ( x ) 3 cot 2 ( x ) cot 4 ( x )   using   cot 2 ( x ) = csc 2 ( x ) 1   =   cot ( x ) csc ( x ) 4 + 5 csc 2 ( x ) csc 4 ( x ) d x + cos 2 ( x ) cos 3 ( x ) 3 sin 2 ( x ) cos ( x ) d x For   the   integrand   cot ( x ) csc ( x ) 4 + 5 csc 2 ( x ) csc 4 ( x )   substitute   u = csc ( x )   and   d u = cot ( x ) csc ( x )    d x   =   1 u 4 + 5 u 2 4 d u + cos 2 ( x ) cos 3 ( x ) 3 sin 2 ( x ) cos ( x ) d x For   the   integrand   1 u 4 + 5 u 2 4   use   partial   fractions:    =   ( 1 6 ( u 1 ) 1 6 ( u + 1 ) + 1 12 ( u + 2 ) 1 12 ( u 2 ) ) d u + cos 2 ( x ) cos 3 ( x ) 3 sin 2 ( x ) cos ( x ) d x Integrate   the   sum   term   by   term   and   factor   out   constants:    =   1 12 1 u + 2 d u + 1 6 1 u + 1 d u 1 6 1 u 1 d u + 1 12 1 u 2 d u + cos 2 ( x ) cos 3 ( x ) 3 sin 2 ( x ) cos ( x ) d x For   the   integrand   1 u + 2   substitute   s = u + 2   and   d s =      d u   =   1 12 1 s d s + 1 6 1 u + 1 d u 1 6 1 u 1 d u + 1 12 1 u 2 d u + cos 2 ( x ) cos 3 ( x ) 3 sin 2 ( x ) cos ( x ) d x The   integral   of   1 s   is   log ( s )   =   log ( s ) 12 + 1 6 1 u + 1 d u 1 6 1 u 1 d u + 1 12 1 u 2 d u + cos 2 ( x ) cos 3 ( x ) 3 sin 2 ( x ) cos ( x ) d x For   the   integrand   1 u + 1   substitute   p = u + 1   and   d p =      d u   =   log ( s ) 12 + 1 6 1 p d p 1 6 1 u 1 d u + 1 12 1 u 2 d u + cos 2 ( x ) cos 3 ( x ) 3 sin 2 ( x ) cos ( x ) d x The   integral   of   1 p   is   log ( p )   =   log ( p ) 6 log ( s ) 12 1 6 1 u 1 d u + 1 12 1 u 2 d u + cos 2 ( x ) cos 3 ( x ) 3 sin 2 ( x ) cos ( x ) d x For   the   integrand   1 u 1   substitute   w = u 1   and   d w =      d u   =   log ( p ) 6 log ( s ) 12 1 6 1 w d w + 1 12 1 u 2 d u + cos 2 ( x ) cos 3 ( x ) 3 sin 2 ( x ) cos ( x ) d x The   integral   of   1 w   is   log ( w )   =   log ( p ) 6 log ( s ) 12 log ( w ) 6 + 1 12 1 u 2 d u + cos 2 ( x ) cos 3 ( x ) 3 sin 2 ( x ) cos ( x ) d x For   the   integrand   1 u 2   substitute   v = u 2   and   d v =      d u   =   log ( p ) 6 log ( s ) 12 log ( w ) 6 + 1 12 1 v d v + cos 2 ( x ) cos 3 ( x ) 3 sin 2 ( x ) cos ( x ) d x The   integral   of   1 v   is   log ( v )   =   log ( p ) 6 log ( s ) 12 + log ( v ) 12 log ( w ) 6 + cos 2 ( x ) cos 3 ( x ) 3 sin 2 ( x ) cos ( x ) d x ⁠  Multiply   numerator   and   denominator   of   cos 2 ( x ) cos 3 ( x ) 3 sin 2 ( x ) cos ( x )   by   csc 2 ( x ) sec ( x )   =   log ( p ) 6 log ( s ) 12 + log ( v ) 12 log ( w ) 6 + cot ( x ) csc ( x ) 3 cot 2 ( x ) d x ⁠  Prepare   to   substitute   z 1 = csc ( x )   Rewrite   cot ( x ) csc ( x ) 3 cot 2 ( x )   using   cot 2 ( x ) = csc 2 ( x ) 1   =   log ( p ) 6 log ( s ) 12 + log ( v ) 12 log ( w ) 6 + cot ( x ) csc ( x ) 4 csc 2 ( x ) d x For   the   integrand   cot ( x ) csc ( x ) 4 csc 2 ( x )   substitute   z 1 = csc ( x )   and   d z 1 = cot ( x ) csc ( x )    d x   =   log ( p ) 6 log ( s ) 12 + log ( v ) 12 log ( w ) 6 + 1 4 z 1 2 d z 1 Factor   4   from   the   denominator:    =   log ( p ) 6 log ( s ) 12 + log ( v ) 12 log ( w ) 6 + 1 4 ( 1 z 1 2 4 ) d z 1 Factor   out   constants:    =   log ( p ) 6 log ( s ) 12 + log ( v ) 12 log ( w ) 6 + 1 4 1 1 z 1 2 4 d z 1 For   the   integrand   1 1 z 1 2 4   substitute   z 2 = z 1 2   and   d z 2 = 1 2    d z 1   =   log ( p ) 6 log ( s ) 12 + log ( v ) 12 log ( w ) 6 + 1 2 1 1 z 2 2 d z 2 The   integral   of   1 1 z 2 2   is   tanh 1 ( z 2 )   =   log ( p ) 6 log ( s ) 12 + log ( v ) 12 log ( w ) 6 + 1 2 tanh 1 ( z 2 ) + constant  Substitute   back   for   z 2 = z 1 2   =   log ( p ) 6 log ( s ) 12 + log ( v ) 12 log ( w ) 6 + 1 2 tanh 1 ( z 1 2 ) + constant  Substitute   back   for   z 1 = csc ( x )   =   log ( p ) 6 log ( s ) 12 + log ( v ) 12 log ( w ) 6 + 1 2 coth 1 ( 2 sin ( x ) ) + constant  Substitute   back   for   v = u 2   =   log ( p ) 6 log ( s ) 12 + 1 12 log ( u 2 ) log ( w ) 6 + 1 2 coth 1 ( 2 sin ( x ) ) + constant  Substitute   back   for   w = u 1   =   log ( p ) 6 log ( s ) 12 + 1 12 log ( u 2 ) 1 6 log ( u 1 ) + 1 2 coth 1 ( 2 sin ( x ) ) + constant  Substitute   back   for   p = u + 1   =   log ( s ) 12 + 1 12 log ( u 2 ) 1 6 log ( u 1 ) + 1 6 log ( u + 1 ) + 1 2 coth 1 ( 2 sin ( x ) ) + constant  Substitute   back   for   s = u + 2   =   1 12 log ( u 2 ) 1 6 log ( u 1 ) + 1 6 log ( u + 1 ) 1 12 log ( u + 2 ) + 1 2 coth 1 ( 2 sin ( x ) ) + constant  Substitute   back   for   u = csc ( x )   =   1 12 log ( csc ( x ) 2 ) 1 6 log ( csc ( x ) 1 ) + 1 6 log ( csc ( x ) + 1 ) 1 12 log ( csc ( x ) + 2 ) + 1 2 coth 1 ( 2 sin ( x ) ) + constant  Factor   the   answer   a   different   way:    =   1 12 ( log ( csc ( x ) 2 ) 2 log ( csc ( x ) 1 ) + 2 log ( csc ( x ) + 1 ) log ( csc ( x ) + 2 ) + 6 coth 1 ( 2 sin ( x ) ) ) + constant  An   alternative   form   of   the   integral   is:    =   1 12 ( log ( csc ( x ) 2 csc ( x ) + 2 ) + 6 coth 1 ( 2 sin ( x ) ) + 4 coth 1 ( csc ( x ) ) ) + constant  方法3. 使用欧拉公式,.作代换.
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所以.
如果利用恒等式化简的话,应该会得出,和上面的结果是相等的.