求定积分问题

求(sinx+tanx)/(cosx+tanx)在0到π/2的积分

matlab运算结果如下，近似值为1.3299。虽然和pi/2+1/4很接近，但不是的。我只想对楼主说卧。槽！浪费我1天时间，就像当初有个逼，让我积一个根本积不出来的不定积分一样。
-1/10*(-10*log(2)*(-2+2*5^(1/2))^(1/2)*(2+2*5^(1/2))^(1/2)+5*log(3-5^(1/2))*(-2+2*5^(1/2))^(1/2)*(2+2*5^(1/2))^(1/2)+log(3-5^(1/2))*5^(1/2)*(-2+2*5^(1/2))^(1/2)*(2+2*5^(1/2))^(1/2)-8*atan(3/(-2+2*5^(1/2))^(1/2)-1/(-2+2*5^(1/2))^(1/2)*5^(1/2))*5^(1/2)*(2+2*5^(1/2))^(1/2)-log(3+5^(1/2))*5^(1/2)*(-2+2*5^(1/2))^(1/2)*(2+2*5^(1/2))^(1/2)+5*log(3+5^(1/2))*(-2+2*5^(1/2))^(1/2)*(2+2*5^(1/2))^(1/2)-8*5^(1/2)*(-2+2*5^(1/2))^(1/2)*Re(atanh((3+5^(1/2))/(2+2*5^(1/2))^(1/2)))-8*5^(1/2)*(2+2*5^(1/2))^(1/2)*atan(-1/(-2+2*5^(1/2))^(1/2)+1/(-2+2*5^(1/2))^(1/2)*5^(1/2))+8*5^(1/2)*(-2+2*5^(1/2))^(1/2)*Re(atanh((5^(1/2)+1)/(2+2*5^(1/2))^(1/2))))/(-2+2*5^(1/2))^(1/2)/(2+2*5^(1/2))^(1/2)

哎，表示无能为力。期待大牛出现。

谢斑竹的分享,

谁有详细步骤？

二分之一π

楠♂ 发表于 2013-4-19 23:06 分子上加个cosx,在减去cosx,就能做出来了

求详细步骤

分子上加个cosx,在减去cosx,就能做出来了