【解析】∫_(-1)^1(x+sinx)dx= (1/2x^2-cosx)|_(-1)^1=1/2-cos1-1/2+cos1=0综上所述,答案是:0.【定积分的概念】一般地,如果函数f(x)在区间[a,b]上连续,用分点a=x_0x_1⋅⋅⋅x_(i-1)x_i⋅⋅⋅x_n=b 将区间 [a,b] 等分成n个小区间,在每个小区间...
答案 【解析】 2 sinxcos2xdx 2 =- cos2xd(cosx) =- (2cos2x-1d(cosx) 入 2 -oskr-cosa) 长o(-)-(-) 2-? 入个 cOS -COS 2 2 =0 T 综上所述,结论:「 sinxcos2xdx=0 2相关推荐 1T2【题目】计算定积分:.T2 反馈 收藏
【题目】计算:∫_0^(π/(2))(cos2x)/(cosx+sinx)dx= 答案 【解析】由题意得,(cos2x)/(cosx+sinx)=(cos^2x-sin^2x)/(cosx+sinx)=cosx-sinx ∴∫_0^(π/2)(cos2x)/(cosx+sinx)dx=∫_0^(π/(2))(cosx-sinx)dx =(sinx+cosx)|_0^2=(1+0)-(0+1)=0综上所述,结论:...
{ 2 } \sin 1 - \frac { \pi ^ { 2 } } { 4 } $$ {(siny-ysinx+2)dx+(cosx+xcosy+x^{2})dy=[C2dx=-π 故{(sin]$$ + \frac { \pi } { 2 } \sin 1 + \frac { \pi ^ { 2 } } { 4 } + $$ (17)取A0:y=0(x:4→0),D为由L+AO所围的平面区域,则 $...
{ \sin 2 x } { \cos x } $$dx;(6)$$ \frac { \sin 2 x } { \sin x } $$dx;(7)$$ \cos ^ { 2 } \frac { x } { 2 } $$dx;$$ 8 ) \int \frac { \cos 2 x } { \cos x - \sin x } d x ; $$(9)$$ \int ( \frac { 1 } { x } - \cos...
错解$$ \int _ { 0 } ^ { \frac { n } { 4 } } \sin 2 x d x = ( - \cos 2 x ) | _ { 0 } ^ { \frac { x } { 4 } } $$ $$ = - \cos \frac { \pi } { 2 } + \cos 0 = 0 + 1 = 1 . $$ 剖析$$ ( - \cos 2 x ) ^ { \prime ...
【解析】解: $$ \int _ { 0 } ^ { \frac { \pi } { 2 } } \sin ^ { 2 } x \cos ^ { 2 } x d x , $$ $$ = \int _ { 0 } ^ { \frac { \pi } { 2 } } \frac { 1 } { 4 } \sin ^ { 2 } 2 x d x , $$ $$ = \int _ { 0 } ^ { \f...
【题目】计算定积分∫_(-π/(2))^(π/(2))cosxcos2xdx 答案 【解析】根据题意,由积化和差公式得cos2xcosx=1/2(cos3x+cosx) ∫_(-π/(2))^(π/(2))cosxcos2xdx 则=∫_-π/2)((cos3x+cosx)/2)dx =1/6sin3x|(π/2+1/2sinx|π/(2))/(π/(2)) =1/6*(-1...
【解析】解∫_(π/4)^(π/4)1/(sinxcosx)dx=2∫_1/(1/t)cse^(x/2)dx=∫_1/2c或∫_1^(1/2)1/(sinxcosx)dx=∫_4/5(dx)/(sin^2x)(cosx)/(sinx)=∫_4/x^ =-∫_1/(^(π/4)(dcost)/(cost)(dx)/(cos)=-[ln1(1/x)]'=1/2ln3相关推荐 1用牛顿-莱布尼茨公式计算下列定...
用分部积分法计算下列定积分:∫_(π/4)^(π/(sin^2x)dx; 答案 解原式=-∫_π/(π/(3))xd(cotx)=-xcotx_(π/(4))^(π/(2))+∫_(π/(6))^(π/(4)) rac(cosx =-(π/3+(√3)/3-π/4+1)+ln|sinx||x/3|=(1/4-(√3)/9)π+ln(√3)/2-ln(√2)/2=(1/4-(...