0=-1？

芮之蓝桃

现求\(\int\tan x\)
即求\(\int\sec x\sin x\)
令\(u=\sec x,\dif v=\sin x\)
则\(\dif u=\sec x\tan x, v=-\cos x\)
易得\(\int\tan x=uv-\int v \dif u\)
即\(\int\tan x=-1-\int-\sec x \sin x\)
即得\(0=-1\)

mathe

不定积分相差一个常数都是正常的，严格书写，等号两边都需要添加一个不同的常数$C_1,C_2$

芮之蓝桃

对于抛物线\(x^2\)=2\(py\),p\(\in\)R我们求其任意段弧长

由弧长公式\(\int\)\(\sqrt{1+(\frac{dy}{dx})^2}\)\(dx\)
则原抛物线弧长\(l\)=\(\int\)\(\sqrt{1+\frac{x^2}{p^2}}\)\(dx\)=\(\frac{1}{p}\)\(\int\)\(\sqrt{p^2+x^2}\)\(dx\)
令x=\(ptanx\)则dx=p\(secx^2\)
则\(l\)=\(\frac{1}{p}\)\(\int\)\(\sqrt{p^2sec^2x}\)\(psec^2x\)\(dx\)
经计算可得\(l\)=\(\frac{px\sqrt{x^2+p^2}+p^3ln|\frac{\sqrt{x^2+p^2}+x}{p}|}{2p^2}\)+C