这道不定积分的答案是多少？还要有过程！
\u6c42\u4e00\u9053\u4e0d\u5b9a\u79ef\u5206\u7684\u89e3\u7b54\u8fc7\u7a0b\u53ea\u8981\u4e24\u6b21\u5206\u90e8\u79ef\u5206\u5c31\u53ef\u4ee5\u7b97\u4e86
\u222bsin3x cosx dx
=sin3x sinx - 3\u222bcos3x sinx dx
=sin3x sinx - 3\u3016-cos3x cosx - 3\u222bsin3x cosx dx\u3017
=sin3x sinx + 3cos3x cosx + 9\u222bsin3x cosx dx
\u79fb\u9879\u5f97
-8\u222bsin3x cosx dx=sin3x sinx + 3cos3x cosx
\u6240\u4ee5
\u222bsin3x cosx dx=-1/8(sin3x sinx + 3cos3x cosx) + c
there are some ready-for-use formulas:
\u222b xⁿ dx xⁿ⁺¹/(n+1) + C
\u222b secxtanx dx = secx + C
\u222b(a\u2192b) ƒ(x) dx = \u222b(a\u2192c) ƒ(x) dx + \u222b(c\u2192b) ƒ(x) dx, where c is between a and b
\u222b |x| dx = x, if x > 0
- x, if x < 0
这是1/4圆的面积公式
几何法，1/4圆的面积
绛旓細鈭玿ln(1+x^2)dx =1/2鈭玪n(1+x^2)dx^2 =1/2鈭玪n(1+x^2)d(1+x^2)=1/2(1+x^2)ln(1+x^2)-1/2鈭(1+x^2)dln(1+x^2)=1/2(1+x^2)ln(1+x^2)-1/2鈭(1+x^2)*1/(1+x^2)d(1+x^2)=1/2(1+x^2)ln(1+x^2)-1/2鈭玠x^2 =1/2(1+x^2)ln(1+x...
绛旓細绠鍗曡绠椾竴涓嬪嵆鍙紝绛旀濡傚浘鎵绀
绛旓細浠nx=t锛岃В涓嶅畾绉垎缁撴灉涓簍^2/2,鍐嶅洖浠nx锛岀粨鏋滀负锛坙nx锛塣2/2
绛旓細涓嶅畾绉垎鐨鍏紡娉
绛旓細浠 u = 鈭歺锛屽垯 x = u^2锛宒x = 2udu锛屽師寮 = 鈭(u^3-1) / (u+1) * 2udu =鈭(2u^3-2u^2+2u-4+4/(u+1))du = 1/2*u^4 - 2/3*u^3 + u^2 - 4u + 4ln|u+1|+C = 1/2*x^2 - 2/3*x^(3/2) + x - 4鈭歺 + 4ln(鈭歺+1) + C ...
绛旓細there are some ready-for-use formulas:鈭 xⁿ dx xⁿ⁺¹/(n+1) + C 鈭 secxtanx dx = secx + C 鈭(a鈫抌) ƒ(x) dx = 鈭(a鈫抍) ƒ(x) dx + 鈭(c鈫抌) ƒ(x) dx, where c is between a and b 鈭 |x| dx = x, if x...
绛旓細妤间富绗竴棰樼‘瀹炵Н閿欎簡锛屼功涓婄殑绛旀閿欎簡涓涓鍙凤紱妤间富鐨勭浜岄閿欎簡涓涓鍙峰拰涓涓郴鏁帮紝涔︿笂鐨勭瓟妗堟槸瀵圭殑銆備笅闈㈠悇鎻愪緵涓夌涓嶅悓鐨勮В娉曪紝瑙佸浘銆
绛旓細鍏堢敤鍒嗛儴绉垎娉曪紝鐒跺悗鍐嶆崲鍏冿紝鍙锛漵int锛岀劧鍚庡氨鑳芥眰鍑烘潵浜嗭紝濡傚浘
绛旓細璁緓=tanu 鍒欏師寮=鈭1/[(1+2tan²u)鈭(1+tan²u)]dx =鈭1/[(1+2tan²u)鈭(1+tan²u)]dx =鈭玞osu/(1+sin²u)du =鈭玠(sinu)/(1+sin²u)=arctan(sinu)+C =arctan(sin(arctanx))+C ...
绛旓細鍥犱负鍏紡 : 鈭玿^a dx = x^(a+1)/(a+1) 鏈 a = 3. 鎵浠ヤ箻浠 1/4
