求∫(x+sinx)/(1+cosx)dx从0到蟺/2的积分 ∫（1+cosx）/(1+cosx-sinx)dx ∫x∧2...
\u6c42(x+sinx)/(1+cosx)\u5728 [0\uff0c\u03c0/2]\u4e0a\u7684\u5b9a\u79ef\u5206(x+sinx)/(1+cosx)\u5728 [0\uff0c\u03c0/2]\u4e0a\u7684\u5b9a\u79ef\u5206\u662f\u03c0/2\u3002
\u222b(x+sinx)/(1+cosx)dx
=\u222b[x+2sin(x/2)cos(x/2)]/[2cos²(x/2)]dx
=\u222b[x/(2cos²(x/2))]dx+\u222b[2sin(x/2)cos(x/2)]/[2cos²(x/2)]dx
=\u222bxdtan(x/2)+\u222btan(x/2)dx
=xtan(x/2)-\u222btan(x/2)dx+\u222btan(x/2)dx
=xtan(x/2)+C
\u6240\u4ee5\u539f\u5b9a\u79ef\u5206
=xtan(x/2)|(0,\u03c0/2)
=\u03c0/2
\u6269\u5c55\u8d44\u6599\uff1a
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\u4e00\u4e2a\u51fd\u6570\uff0c\u53ef\u4ee5\u5b58\u5728\u4e0d\u5b9a\u79ef\u5206\uff0c\u800c\u4e0d\u5b58\u5728\u5b9a\u79ef\u5206\uff1b\u4e5f\u53ef\u4ee5\u5b58\u5728\u5b9a\u79ef\u5206\uff0c\u800c\u4e0d\u5b58\u5728\u4e0d\u5b9a\u79ef\u5206\u3002\u4e00\u4e2a\u8fde\u7eed\u51fd\u6570\uff0c\u4e00\u5b9a\u5b58\u5728\u5b9a\u79ef\u5206\u548c\u4e0d\u5b9a\u79ef\u5206\uff1b\u82e5\u53ea\u6709\u6709\u9650\u4e2a\u95f4\u65ad\u70b9\uff0c\u5219\u5b9a\u79ef\u5206\u5b58\u5728\uff1b\u82e5\u6709\u8df3\u8dc3\u95f4\u65ad\u70b9\uff0c\u5219\u539f\u51fd\u6570\u4e00\u5b9a\u4e0d\u5b58\u5728\uff0c\u5373\u4e0d\u5b9a\u79ef\u5206\u4e00\u5b9a\u4e0d\u5b58\u5728\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
\u5982\u56fe\u6240\u793a\uff1a\u7b2c\u4e00\u9898\uff1a
\u7b2c\u4e8c\u9898\uff1a
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绛旓細鈭玿sinx dx 鍒╃敤鍒嗛儴绉垎娉 =-鈭玿dcosx =-xcosx+鈭玞osx dx =-xcosx+sinx+c
