三角函數及反三角函數之積分 - 銘傳大學-銘傳網頁
1 cos 2 sin dx x I xdx ò = - p p 0 cos 2 2 1 2 xdx [ ] p p 0 sin 2 4 1 2 = - x (0 0 ) 4 1 2 = - - p 2 p = 例 2. (解 ) Find ò xdx 4 sin ò I = xdx 4 sin ò ò - = = dx x x dx 2 2 2 ) 2 1 cos 2 (sin ) ( ò = ( 1 - 2 cos 2 x + cos 2 x ) dx 4 1 2 x xdx C x = - + + ...