arctant-1等于多少值

鬱叙昆3639计算不定积分∫arctan√xdx -
靳涛泰19867016149 ______[答案] √x=tx=t²dx=2tdt∫arctan√xdx=∫2tarctantdt=∫arctantdt²=t²arctant-∫t²/(1+t²)dt=t²arctant-∫(1+t²-1)/(1+t²)dt=t²arctant-∫1-1/(1+t²)dt=t²arctant...

鬱叙昆3639(x+y)dx+(y - x)dy=0,y(1)=0 -
靳涛泰19867016149 ______[答案] ∵令y=xt,则dy=xdt+tdx 代入原方程,化简得 (1-t)dt/(1+t^2)=dx/x ==>arctant-(1/2)ln(1+t^2)=ln│x│+ln│C│ (C是常数) ==>arctant=(1/2)ln(1+t^2)+ln│x│+ln│C│ ==>e^(arctant)=Cx√(1+t^2) ==>e^(arctan(y/x))=C√(x^2+y^2) ∴原方程的通解是e^...

鬱叙昆3639求不定积分arctan(1/x)/(1+x2)dx -
靳涛泰19867016149 ______[答案] 令1/x = t 则原式= ∫arctant/(1 + 1/t²) * (-1/t²)dt =∫-arctant/(1+t²) dt =∫-arctant darctant =-1/2 arctan²t + C =-1/2 arctan(1/x²) + C

鬱叙昆3639已知tanx=1/t,那么arctant等于多少 -
靳涛泰19867016149 ______[答案] tanx=1/t t=1/tanx=cotx arctant=arctan(cotx) 然后证明arctan(cotx)=π/2-x 令f(x)=arctan(cotx)-π/2+x 显然f(0)=0 f'(x)=1/(cot^2(x)+1)*(-1/sin^2(x))+1=-1/(sin^2(x)+cos^2(x))+1=0 所以f(x)=0恒成立,即arctan(cotx)=π/2-x

鬱叙昆3639求函数f(x)=∫(上限x,下限0)(t+1)arctant dt 的极值 -
靳涛泰19867016149 ______ 求函数f(x)=(0,x)∫(t+1)arctant dt 的极值 解:令df(x)/dx=(x+1)arctanx=0 得驻点x₁=-1,x₂=0 为书写简便,先求不定积分.∫(t+1)arctantdt=∫t(arctant)dt+∫arctantdt 其中∫arctantdt=t(arctant)-ln[√(1+t²)] ∫tarctantdt=(1/2)∫arctantd(t²)=(1/2){t...

鬱叙昆3639arctan((1 - x^2)/(1+x^2))的不定积分是什么? -
靳涛泰19867016149 ______ ^^^∫5261x^2arctanx/(1+x^41022)dx =∫1653(x^2+1-1)arctanx/(1+x^2)dx =∫arctanxdx-∫arctanx/(1+x^2)dx =xarctanx-∫ x/(1+x^2)dx-∫arctanxd (arctanx) =xarctanx-(1/2)∫ 1/(1+x^2)d(x^2)-(1/2)(arctanx)^2 =xarctanx-(1/2)ln(1+x^2)-(1/2)(arctanx)^2+C

鬱叙昆36392t - arctant|(上限1,下限0)怎么计算? -
靳涛泰19867016149 ______ ∫2tdt=t^2 ∫arctantdt=t*arctant-∫td(arctant) =t*arctant-∫t/(1+t^2)dt =t*arctant-(1/2)∫1/(1+t^2)d(t^2+1) =t*arctant-(1/2)*ln(1+t^2) 所以原式=t^2-t*arctant+(1/2)*ln(1+t^2) t=1,t^2-t*arctant+(1/2)*ln(1+t^2)=1-π/4+1/2*ln2 t=0,t^2-t*arctant+(1/2)*ln(1+t^2)=0 所以原式=1-π/4+1/2*ln2

鬱叙昆3639∫10arctan(1+√x)dx的解是多少?很急,很急,谢谢 -
靳涛泰19867016149 ______ 令1+根号x=t ∫arctantd(t-1)^2=arctant*(t-1)^2-∫(t-1)^2darctant =arctant*(t-1)^2-∫(t-1)^2/(1+t^2)dt =arctant*(t-1)^2-∫(1-2t/(1+t^2))dt =arctant*(t-1)^2-(t-ln|1+t^2|)+C =arctant*(t-1)^2+ln|1+t^2|-t+C 代入t=根号x+1 即可 楼主应该会把 望采纳 谢谢

鬱叙昆3639当u足够大时,arctanu>1,所以∫n到2n arctanudu>2n - n =n ,为什么>2n - n -
靳涛泰19867016149 ______ 按照积分中值定理,必然存在一个数t属于(n,2n),使得arctant*(2t-t)=tarctant.当u趋于无穷大时,t趋于无穷大,因此arctant=pi/2...

鬱叙昆3639证明,当x>1时,有xlnx>arctan(x - 1) -
靳涛泰19867016149 ______[答案] 令x-1=t>0, x=t+1 即证: (t+1)ln(t+1)>arctant 令f(t)=(t+1)ln(t+1)-arctant f'(t)=ln(t+1)+1-1/(1+t平方) =ln(t+1)+t平方/(1+t平方)>0 所以 当t>0时,f(t)是增函数, 从而 f(t)>f(0)=0 即 (t+1)ln(t+1)-arctant>0 (t+1)ln(t+1)>arctant 亦即 xlnx>arctan(x-1), (x>1)