∫sinx+cosx分之cos2xdx怎么算

2025-03-16 07:03:35
推荐回答(1个)
回答1:

∫cosx/(sinx+cosx) dx
= (1/2)∫[(cosx+sinx)+(cosx-sinx)]/(sinx+cos)]dx
= (1/2)∫ dx + (1/2)∫(cosx-sinx)/(sinx+cosx)dx
= x/2 + (1/2)∫d(sinx+cosx)/(sinx+cosx)
= (1/2)(x+ln|sinx+cosx|) + C(C为常数)
扩展资料:
不定积分求法:
1、积分公式法。直接利用积分公式求出不定积分。
2、换元积分法。换元积分法可分为第一类换元法与第二类换元法。
(1)第一类换元法(即凑微分法)。通过凑微分,最后依托于某个积分公式。进而求得原不定积分。
(2)第二类换元法经常用于消去被积函数中的根式。当被积函数是次数很高的二项式的时候,为了避免繁琐的展开式,有时也可以使用第二类换元法求解。
3、分部积分法。设函数和u,v具有连续导数,则d(uv)=udv+vdu。移项得到udv=d(uv)-vdu
两边积分,得分部积分公式∫udv=uv-∫vdu。
常用不定积分公式
1、∫kdx=kx+C。
2、∫x^ndx=[1/(n+1)]x^(n+1)+C。
3、∫sinxdx=-cosx+C。
4、∫cosxdx=sinx+C。

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