企源知识库

求下列定积分：
(1)∫₀¹xcosπxdx；
(2)∫₀^√2/2arccosxdx；
(3)∫₀¹xarctanxdx
(4)∫₀¹x²e^x/2dx；
(5)∫₀¹xln(1+x²)dx；
(6)∫₁^e(lnx/x³)dx
(7)∫_π/4^π/2(x/sin²x)dx；
(8)∫₀^πe^-xsinxdx；
(9)∫_e^1/e|lnx|dx(10)∫_-1/2^1/2(xarcsinx/√1-x²)dx；
(11)已知xe^x为f(x)的一个原函数，求∫₀¹xf′(x)dx．
【正确答案】：(1)∫₀¹xcosπxdx=1/π∫₀¹(sinπx) =1/π[xsinx|₀¹-∫₀¹sinπxdx] =(1/π²)cosπx|₀¹=-(2/π²) (2)∫₀^√2/2arccosxdx=xarccosxdx=xarccos|₀^√2/2-∫₀^√2/2x•-(1/√1-x²)dx =xarccosx|₀^√2/2-1/2∫₀^√2/2[d(1-x²)/√1-x² =xarccosx|₀^√2/2-√1-x²|₀^√2/2 =(√2/8)π-√2/2+1 (3)∫₀¹xarctanxdx=(1/2)∫₀¹arctannxdx² =1/2[x²arctanx|₀¹-∫₀¹x²/(1+x²)dx] =1/2[π/4-(x-arctanx)|₀¹ =π/4-1/2 (4)∫₀¹x²e^x/2dx =2∫₀¹x²de^x/2 =2[x²e^x/2|₀¹-∫₀¹e^x/2•2xdx] =2[e^1/2-4∫₀¹xde^π/2] =2[e^1/2-4xe^x/2|₀¹+4∫₀¹e^x/2dx] =10e^1/2-16 (5)∫₀¹xln(1+x²)dx=(1/2)∫₀¹ln(1+x²)d(x²+1) =[(x²+1)/2]ln(1+x²)|₀¹-1/2∫₀¹[2x(x²+1)/(x²)+1]dx =ln2-x²/2|₀¹=ln2-1/2 (6)∫₁^e(lnx/x³)dx=-(1/2)∫_e¹lnxd(1/x)² =-(1/2)[1/x²lnx|₁^e-∫₁^e1/x²•(1/x)dx] =-(1/2)[1/e²+1/2x²|₁^e] =1/4-3/4e² (7)∫_π/4^π/2(x/sin²x)dx=-∫_π/4^π/2xdcotx|_π/4^π/2+∫_π/4^π/2cotxdx =π/4+(1/2)ln2 (8)∫₀^πe^-xsinxdx =-∫₀^πsinde^-x =-sinx•e^-x|₀^π+∫₀^πe^-xcosxdx =-∫₀^πcosxde^-x =-e^-xcosx|₀^π+∫₀^πe^-x(-sinx)dx 所以 2₀^πe^-xsinxdx=-e^-xcosx|₀^π=e^-π+1 所以 ∫₀^πe^xsinxdx=(e^-π+1)/2 (9)∫_1/e^e|lnx|dx =-_1/e¹lnxdx+_e^elnxdx =-xlnx|_1/e¹+∫_1/e¹xdlnx+xlnx-|_e¹-∫₁^exdlnx =-(1/e)+x|_1/e¹+e-x|₁^e =2(1-e^-1) (10)∫_-(1/2)^1/2(xarcsinx/√1-x²)dx =-∫₀^1/2(arcsinx/√1-x²)d(1-x²) =-2∫₀^1/2arcsinxd√1-x² =-2√1-x²arsinx+|₀^1/2+2∫₀^1/2√1-x²darcsinx =-(√3/6)π-2x|₀^1/2 =-(√3/6)π+1 (11)由题意知f(x)=e^x+xe^x 故∫₀¹xf′(x)dx=∫₀¹xdf(x)|₀¹-∫₀¹f(x)dx =xf(x)|₀¹-xe^x|∫₀¹ =2e-e =e