=e xsinx e xcosx ∫e xcosxdx,
11所以 ∫e xcosxdx=(e xsinx e xcosx)+C=e x(sinx cosx)+C. 22
x 7. ∫e 2xsindx; 2
解 因为
xxxx ∫e 2xsindx=2∫e 2xdcos=2e 2xcos 2∫cosde 2x 2222
xxxx =2e 2xcos+4∫e 2xcosdx=2e 2xcos+8∫e 2xdsin 2222
xxx =2e 2xcos+8e 2xsin 8∫sinde 2x 222
xxx =2e 2xcos+8e 2xsin+16∫e 2xsindx, 222
x2xx所以 ∫e 2xsindx= e 2x(cos+4sin+C. 21722
x 8. ∫xcosdx; 2
xxxxxx 解 ∫xcosdx=2∫xdsin=2xsin 2∫sindx=2xsin+4cos+C. 222222
9. ∫x2arctanxdx;
解 ∫x2arctanxdx=1131313xdx=xx x dx arctanarctan3∫33∫1+x2
131x211213 =xarctanx ∫dxxarctanx(1dx2 = ∫22361+x361+x
111 =x3arctanx x2+ln(1+x2)+C. 663
10. ∫xtan2xdx
1 解 ∫xtan2xdx=∫x(sec2x 1)dx=∫xsec2xdx ∫xdx= x2+∫xdtanx 2
11 = x2+xtanx ∫tanxdx= x2+xtanx+ln|cosx|+C. 22
11. ∫x2cosxdx;
解 ∫x2cosxdx=∫x2dsinx=x2sinx ∫sinx 2xdx=x2sinx+2∫xdcosx