①∫ tan√(1+x^2)*x/√(1-x^2)dx ②∫f'(arcsinx)*1/√(1-x^2)dx①∫ tan√(1+x^2)*x/√(1-x^2)dx②∫f'(arcsinx)*1/√(1-x^2)dx

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①∫ tan√(1+x^2)*x/√(1-x^2)dx ②∫f'(arcsinx)*1/√(1-x^2)dx①∫ tan√(1+x^2)*x/√(1-x^2)dx②∫f'(arcsinx)*1/√(1-x^2)dx
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①∫ tan√(1+x^2)*x/√(1-x^2)dx ②∫f'(arcsinx)*1/√(1-x^2)dx①∫ tan√(1+x^2)*x/√(1-x^2)dx②∫f'(arcsinx)*1/√(1-x^2)dx
①∫ tan√(1+x^2)*x/√(1-x^2)dx ②∫f'(arcsinx)*1/√(1-x^2)dx
①∫ tan√(1+x^2)*x/√(1-x^2)dx
②∫f'(arcsinx)*1/√(1-x^2)dx

①∫ tan√(1+x^2)*x/√(1-x^2)dx ②∫f'(arcsinx)*1/√(1-x^2)dx①∫ tan√(1+x^2)*x/√(1-x^2)dx②∫f'(arcsinx)*1/√(1-x^2)dx
∫tan(√(1+x^2) xdx/√(1+x^2)
=∫tan√(1+x^2)d √(1+x^2)
=-ln|cos√(1+x^2)|+C
∫f'(arcsinx)dx/√(1-x^2)
=∫f'(arcsinx)d(arsinx)
=f(arcsinx)+C

∫sec^2 x / √ tan x +1 dx ∫tan√(1+x^2)*x/(√1+x^2)dx 求证tan(x+y)*tan(x-y)=tan^2x-tan^2y/1-tan^2xtan^2y tan（ x/2+π/4）+tan（x/2-π/4 ）=2tanxtan(x/2+π/4)+tan(x/2-π/4)=[tan(x/2)+tan(π/4)]/[1-tan(x/2)tan(π/4)]+[tan(x/2)-tan(π/4)]/[1+tan(x/2)tan(π/4)]=[tan(x/2)+1]/[1-tan(x/2)]+[tan(x/2)-1]/[1+tan(x/2)]=[(tan(x/2)+1)^2-(tan(x/2)-1)^2]/[1-(tan(x (tan x+1/tan)cos^2 x等于? ￡tan x(tan x+1)dx 化简sin x/(1－cos x)√[(tan x－sin x)（tan）]化简sin x/(1－cos x)√[(tan x－sin x)/（tan x＋sin x）] 数学 [tan(x/2) +1 ]/[1 - tan(x/2)] + [tan(x/2) - 1]/[1 + tan(x/2)] 怎麼化简? tan^2x+1等于? tan^-1(x^2) .求导! 1、∫x√x^2+1dx2、∫e^xsin(e^x-2)dx3、∫(lnx/x)dx4、∫(1/x^2)乘tan(1/x)dx5、∫cos^2x乘sinxdx6、∫tan^5xsec^2xdx 极限(1-x)tanπx/2 (1+tan(x))/sin(2x)不定积分 x(tan^-1 x)^2积分 化简1/cos^2x*√（1+tan^2x）(x为第二象限角) 求∫1／tan^2x+sin^2x dx ∫e^x*tan(x-1)^2dx 这道题为什么第二步的+到了第三步变成了-1.tan(x/2+π/4)+tan(x/2-π/4)=[tan(x/2)+tan(π/4)]/[1-tan(x/2)tan(π/4)]+[tan(x/2)-tan(π/4)]/[1+tan(x/2)tan(π/4)]=[tan(x/2)+1]/[1-tan(x/2)]+[tan(x/2)-1]/[1+tan(x/2)]=[(tan(x/2)+1)^2-(tan(x/2)-