评估以下积分。
问题11∫EX(sin4x-4)/(2sin 2 2×)DX
解决方案:
We have,
∫ex(sin4x-4)/(2sin22x)dx
=∫ex(2sin2xcos2x-4)/(2sin22x)dx
=∫ex(((2sin2xcos2x)/(2sin22x))-4/(2sin22x))dx
=∫ex(cot2x-2cosec22x)dx
=∫excot2xdx-2∫excosec22xdx
Integrating by parts,
excot2x-2∫exd(cot2x)/dx-2∫excosec22xdx
= excot2x+2∫excosec22xdx-2∫excosec22xdx
= excot2x+C
问题12∫EX(2-X)/(1-X)2 DX
We have,
∫ex(2-x)/(1-x)2dx
=∫ex((1-x)+1)/(1-x)2dx
=∫ex(((1/(1-x))+(1/(1-x)2)))dx
=∫ex(1/(1-x))+∫ex(1/(1-x)2)dx
= ex/(1-x)+C
问题13∫E×(1 + X)/(X + 2)2 DX
We have,
∫ex(1+x)/(x+2)2dx
=∫ex((2+x)-1)/(x+2)2dx
=∫ex(((2+x)/(x+2)2)-1/(x+2)2)dx
=∫ex((1/(x+2))-1/(x+2)2)dx
= ex/(x+2)dx
问题14.∫e (-x / 2) (1-sinx) (1/2) /(1 + cosx)dx
We have,
∫e(-x/2)(1-sinx)(1/2)/(1+cosx)dx
Let x/2 = t
So, x = 2t
So the equation is,
∫2e(-t)(1-sin2t)(1/2)/(1+cos2t)dt
=2∫e(-t)(sin2t+cos2t-2sintcost)(1/2)/(2cos2(t))dt
=∫e(-t)(sint-cost)(2*1/2)/cos2tdt
=∫e(-t)(sint-cost)/cos2tdt
=∫e(-t)(tantsect-sect)dt
=∫e(-t)(tant sect)dt-∫e(-t)sectdt
=∫e(-t)(tant sect)dt -e(-t)sect -∫e(-t)(d(sect)/dt) dt
=∫e(-t)(tant sect)dt -e(-t)sect -∫e(-t)sect tantdt
= e(-t)sect
= e(-x/2)sec(x/2)+C
问题15.∫EX(logX的+ 1 / x)的DX
解决方案:
We have,
∫ex(logx +1/x)dx
= ex(logx)+C
问题16∫EX(logX的+ 1 /×2)DX
解决方案:
We have,
∫ex(logx +1/x2)dx
=∫ex(logx+1/x-1/x+1/x2)dx
=∫ex((logx-1/x)+((1/x)+(1/x2))dx
= ex(logx-1/x)+C
问题17∫EX / X(X(logX的)2 + 2logx)DX
解决方案:
We have,
∫ex/x(x(logx)2+2logx)dx
=∫ex((logx)2+2ex(logx)/x)dx
=∫ex(logx)2dx +∫(2ex/x)(logx)dx
Integrating by parts,
= ex(logx)2-∫ex(d(logx)2/dx)dx +∫(2ex/x)(logx)dx
= ex(logx)2-∫(ex/x)2logxdx+∫2ex/x(logx)dx
= ex(logx)2+C
问题18∫EX(-1罪X + 1 /(1-X 2)1/2)DX
解决方案:
= exsin-1x-∫ex(d(sin-1x)/dx) dx +∫ex/(1-x2)1/2dx
= exsin-1x- ∫ex/(1-x2)1/2dx+∫ex/(1-x2)1/2dx
= exsin-1x+C
问题19∫E2×(-sinx + 2cosx)DX
解决方案:
= -∫e2xsinxdx +2∫e2xcosxdx
= -∫e2xsinxdx+2((1/2)e2xcosx+∫(1/2)e2xsinxdx)
= -∫e2xsinxdx+e2xcosx+∫e2xsinxdx
= e2xcosx+C
问题20∫EX(TAN-1 X + 1 /(1 + X 2))DX
解决方案:
= extan-1x-∫ex (d(tan-1x)/dx) dx+∫ex/(1+x2)dx
= extan-1x-∫ex/(1+x2)dx+∫ex/(1+x2)dx
= extan-1x+C
问题21∫E×((sinxcosx-1)/ SIN 2×)DX
解决方案:
= ∫ex(cotx-cosec2x)dx
= excotx-∫ex(d(cotx)/dx) dx-∫excosec2xdx
= excotx+∫excosec2xdx -∫excosec2xdx
= excotx+C
问题22.∫(tan(logx)+ sec 2 (logx))dx
解决方案:
Suppose,
logx=z
=>ez=x
=>d(ez)/dx=1
=>ezdz=dx
Substituting it in original question,
∫(tan z+sec2z)ezdz
= eztanz -∫ez(d(tanz)/dz))dz+∫ezsec2zdz
= eztanz-∫ezsec2zdz+∫ezsec2zdz
= eztanz+C
= e(logx)tan(logx)+C
= x tan(logx)+C
问题23∫EX(X-4)/(X-2)3 DX
解决方案:
= ∫ex((x-2)-2)/(x-2)3dx
= ∫ex((1/(x-2)2)-(2/(x-2)3))dx
= ex/(x-2)2-∫ex(d((x-2)-2)/dx)dx-2∫ex/(x-2)3dx
= ex/(x-2)2+2∫ex/(x-2)3dx -2∫ex/(x-2)3dx
= ex/(x-2)2+C
问题24∫E2×((1-sin2x)/(1-2cosx))DX
解决方案:
=∫e2x((1-sin2x)/2sin2x)dx
=∫e2x((cosec2x/2)-cotx)dx
Suppose,
I=I1+I2
I1=1/2∫e2xcosec2xdx
I2=-∫e2xcotxdx
let,
u=e2x
=du=2e2xdx
and
∫cosec2xdx=∫dv
=>v=-cotx+C
So,
I1=1/2[e2x(-cotx)-∫(-cotx)2e2xdx]
I1=1/2(e2x(-cotx))+∫cotxe2xdx
Thus,
I=(1/2)(e2x(-cotx))+∫cotxe2xdx -∫e2x cotx dx
=>I=(1/2)(e2x(-cotx))+C