快考试了，不想挂科。求(sinx)^2 / (cosx)^2的不定积分，谢谢
\u6c42\u4e0d\u5b9a\u79ef\u5206(sinx*cosx)/(sinx+cosx)\u89e3\uff1a
\u539f\u5f0f=\u222b[(sinx+cosx)^2-1]/2(sinx+cosx)dx
=(1/2)\u222b[(sinx+cosx)-1/(sinx+cosx)]dx
=(1/2)\u222b(sinx+cosx)dx-(1/2)\u222b1/(sinx+cosx)dx
\u7531\u4e8e(sinx+cosx)\u53ef\u5316\u4e3a\u6839\u53f72*sin(x+\u03c0/4)\u2026\u2026\u2026\u2026\u89e3\u91ca\uff1a\u03c0\u4e3a\u5706\u5468\u7387\uff0c\u53733.14159\u2026\u2026\u6240\u4ee5\uff1a
=(1/2)*(sinx-cosx)-(1/2\u6839\u53f72)ln[((\u6839\u53f72)-cosx+sinx)/(sinx+cosx)]+c
\u7531\u4e8e\u65b9\u6cd5\u7684\u4e0d\u540c\uff0c\u7b54\u6848\u4e5f\u4f1a\u4e0d\u4e00\u6837\uff0c\u60a8\u53ef\u4ee5\u9a8c\u8bc1\u4e00\u4e0b\u6211\u7684\u65b9\u6cd5\uff0c\u5982\u679c\u548c\u60a8\u7684\u7ed3\u679c\u4e00\u81f4\uff0c\u7ed9\u70b9\u8f9b\u82e6\u5206\u5427\uff0c\u5475\u5475\uff01
\u222bcosx/(sinx+cosx) dx=(1/2)(x+ln|sinx+cosx|) + C\u3002\uff08C\u4e3a\u79ef\u5206\u5e38\u6570\uff09
\u89e3\u7b54\u8fc7\u7a0b\u5982\u4e0b\uff1a
\u222bcosx/(sinx+cosx) dx
= (1/2)\u222b[(cosx+sinx)+(cosx-sinx)]/(sinx+cos)] dx
= (1/2)\u222b dx + (1/2)\u222b(cosx-sinx)/(sinx+cosx) dx
= x/2 + (1/2)\u222bd(sinx+cosx)/(sinx+cosx)
= (1/2)(x+ln|sinx+cosx|) + C
\u6269\u5c55\u8d44\u6599\uff1a
\u5206\u90e8\u79ef\u5206\uff1a
(uv)'=u'v+uv'\uff0c\u5f97\uff1au'v=(uv)'-uv'\u3002
\u4e24\u8fb9\u79ef\u5206\u5f97\uff1a\u222b u'v dx=\u222b (uv)' dx - \u222b uv' dx\u3002
\u5373\uff1a\u222b u'v dx = uv - \u222b uv' d,\u8fd9\u5c31\u662f\u5206\u90e8\u79ef\u5206\u516c\u5f0f\u3002
\u4e5f\u53ef\u7b80\u5199\u4e3a\uff1a\u222b v du = uv - \u222b u dv\u3002
\u5e38\u7528\u79ef\u5206\u516c\u5f0f\uff1a
1\uff09\u222b0dx=c
2\uff09\u222bx^udx=(x^(u+1))/(u+1)+c
3\uff09\u222b1/xdx=ln|x|+c
4\uff09\u222ba^xdx=(a^x)/lna+c
5\uff09\u222be^xdx=e^x+c
6\uff09\u222bsinxdx=-cosx+c
7\uff09\u222bcosxdx=sinx+c
8\uff09\u222b1/(cosx)^2dx=tanx+c
9\uff09\u222b1/(sinx)^2dx=-cotx+c
10\uff09\u222b1/\u221a\uff081-x^2) dx=arcsinx+c
11\uff09\u222b1/(1+x^2)dx=arctanx+c
先用公式化成对cos2x的不定积分,再把cos2x^用cos4x表示,就变成对一次的积分了,就好算了
绛旓細(sinx)^2 / (cosx)^2锛(tanx)^2锛(secx)^2锛1锛屽鐢ㄤ笉瀹氱Н鍒嗗叕寮忥紝缁撴灉鏄痶anx锛峹锛婥
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