\int (x+1)^2dx=\int x^2+2x+1dx=\frac{1}{3}x^3+x^2+x+C 法二:换元法令t=x+1 , \frac{dt}{dx}=1\Rightarrow dt=dx,把所有的 x 都用t 来代替,于是 \int (x+1)^2dx=\int t^2 dt=\frac{1}{3}t^3=\frac{1}{3}(x+1)^3+C。 我们可以发现通过代换这样积分就简单了很多...
(可做答案-|||-=1/2ln|csc2x-cot2x|+c ()-|||-方法二:凑微分法-|||-∫(dx)/(sin2x)=∫(dx)/(2sinxcosx)=∫(cosxdx)/(2sinxcos^2x) -|||-cos xdx-|||-7-|||-2sin xcos2x-|||-=-1/2∫(dx)/(tanxcos^2x)=-1/2∫(dtanx)/(tanx)-|||-=1/2ln|tanx|+c (-|||-(...
所以 ∫e^xcos2xdx=e^3(3/(13)cos2x+2/(13)sin2x)+( . (10) ∫xcot^2xdx=∫_x^1x(csc^2x-1)dx=-∫xdcotx-∫xdx =-(xc01.x-∫cotxdx)-(x^2)/2 =-xcolx+ln|sinx|-(x^2)/2+C (11) ∫arcsinxdx=xπrcsinx-∫(√(1-x^2))dx=xarcsinx+1/2∫1/(√(1-x^...
∫dx/sin2x =∫(sin^2 x +cos^2 x)dx/2sinxcosx =1/2∫sinxdx/cosx +1/2∫cosxdx/sinx =-1/2∫dcosx/cosx +1/2∫dsinx/sinx =-1/2lncosx +1/2lnsinx +C =1/2ln(sinx/cosx)+C =ln√(sinx/cosx) +C 答案如图所示。
\eqalign{ & \int {\frac{{2x - 3}}{{{x^2} - 3x + 1}}{\kern 1pt} {\text{d}}x} \cr & u = {x^2} - 3x + 1,{\text{d}}x = \frac{1}{{2x - 3}}{\kern 1pt} {\text{d}}u, \cr & = \int {\frac{1}{u}{\kern 1pt} {\text{d}}u} \cr & = \ln u ...
\\ 82.\text{求极限:}L_{82}=\lim_{x\rightarrow 0} \frac{\csc x-\cot x}{x}. \\ 83.\text{设}f\left( x \right) =\frac{2ax^2-\left( a-2 \right) x-1}{ax^2-\left( a^2-1 \right) x-a}\text{;} \\ \left( 1 \right) \text{当}a\text{为何值时,}\lim_{x\...
解原式 =∫x⋅1/(sin^3x)dsinx=-1/2∫xd(1/(sin^2x)) =-1/2[x/(sin^2x)-∫1/(sin^2x)dx]=-1/2xcos^2x+1/2∫cose^2xdx =-1/2xcsc^2x-1/2cotx+c . 解原式 =∫x⋅1/((1+x^2)^2)dx=1/2∫x⋅1/((1+x^2)^2)d(1+x^2) =-1/2∫xd(1/(1+x^2))...
=tanx/2+ln|∈e^(x/2+C)=tanx/2+ln(1+tan^2x/2)+C (9) ∫(sinx)/(sin^2x+5cos^2x)dx=∫(d(-cosx))/(1+4cos^2x)=-1/2∫(d(2cosx))/(1+4cos^2x)=-\frac( . ,) (10) ) ∫(1+tanx)/(sin2x)dx=∫(1/(sin2x))+1/(2cos^2x)dx=1/2ln|csc2x-c(12x)+1/2ta...
= ∫ csc2x dx + ∫ csc2x * tanx dx = (1/2)∫ csc2x d(2x) + ∫ sinx/cosx * 1/(2sinxcosx) dx = (1/2)ln|ducsc(2x)-cot(2x)| + (1/2)∫ sec²x dx = (1/2)ln|1/sin2x-cos2x/sin2x| + (1/2)tanx + C = (1/2)ln|[1-(1-2sin²x)]/(...
方法如下，请作参考：