I=\int \frac{1}{\cos x(1+\sin x)}dx=\frac12\int\Big[\frac1{\cos x}+\frac{\cos x}{(1+\sin x)^2}\Big]dx =\frac12\Big[\ln|{\sec x}+\tan x|-\frac{1}{1+\sin x}\Big]+C. 方法二： I=\int \frac{1}{\cos x(1+\sin x)}dx=\int\sec xd\Big(\frac{-\cos x}{1+\sin x}\Big) =-\frac1{1+\sin x}+\int\frac{\cos x}{1+\sin …

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∫1/cosxdx∫1/cosxdx=∫ cosx/cos²xdx=∫ 1/(1-sin²x) d(sinx)=(1/2)∫ [1/(1+sinx)+1/(1-sinx)] d(sinx)=(1/2) [ln(1+sinx)-ln(1-sinx)] + C=ln √[(1+sinx)/(1-si

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d(f(x))=f'(x)dx. 所以d(cos(x))=cos'(x)dx=-sin(x)dx. dy/dx是我们通常理解的导数，表示对y关于x求导，dx是由此引发的微分算子

Dec 29, 2019 · dsinx和cosxdx分别表示什么，dsinx=cosxdx这个等式的解释。 解：dx表示微分，也称作微元因为sinx的导数是cosx，dsinx相当于sinx对x微分，即d (sinx)= (sinx)'dx=cosxdx.通常把自变量x的增量Δx称为自变量的微分，记.

同样用倍角公式消去平方： \begin {aligned} \int sin^2x~dx&=\int \frac {1-cos2x} {2}~dx=\frac {1} {2}x-\frac {1} {4}sin2x+C \end {aligned} \\\begin {aligned} \int sin^2x~dx&=\int \frac {1-cos2x} {2}~dx=\frac {1} {2}x-\frac {1} {4}sin2x+C \end {aligned}\\

结论一. \color{red}{\int_{0}^{\frac{\pi}{2}}f(\sin x)dx=\int_{0}^{\frac{\pi}{2}}f(\cos x)dx} 证 \int_{0}^{\frac{\pi}{2}}f(\sin x)dx ,令x=\frac{\pi}{2}-t\Rightarrow \sin x=\cos t,dx=-dt

May 7, 2019 · 对于含有正弦函数sin(x)、余弦函数cos(x)、正切函数tan(x)等三角函数的积分，通常需要用到三角恒等变换或者三角积分表。 文档中会详细说明这些积分的 推导 过程。

\frac{d}{dx}(\frac{3x+9}{2-x}) (\sin^2(\theta))' \sin(120) \lim _{x\to 0}(x\ln (x)) \int e^x\cos (x)dx \int_{0}^{\pi}\sin(x)dx \sum_{n=0}^{\infty}\frac{3}{2^n} 显示更多