积分计算题求解
1.设x=1/t
∫桥如旦1/x(x^6+1)dx
=∫(1/[1/t(1/t^6+1)]d(1/t)
=∫(t^7*(-1/t^2)/(1+t^6)dt
=-∫t^5/(1+t^6)dt
=-1/6∫1/(1+t^6)d(1+t^6)
=-1/6ln|1+t^6|+C
将t=1/x代橡枣入,原式=-1/6ln|1+1/x^6|+c
2.先利用积化和差公式:
sin(A)cos(B) = (1/2)[sin(A+B) + sin(A-B)]
∴sin(2x)cos(3x) = (1/2)[sin(2x+3x) + sin(2x-3x)]
= (1/2)[sin5x + sin(-x)]
= (1/2)(sin5x - sinx)
∴∫ sin(2x)cos(3x) dx
= (1/2)∫ sin(5x) dx - (1/2)∫ sinx dx
= (1/2)(1/5)∫ sin(5x) d(5x) - (1/2)∫ sinx dx
= (1/10)(-cos(5x)] + (1/2)cosx + C
= (1/10)[5cosx - cos(5x)] + C
4.=∫敏扰(1/x^4-1/x^6)dx=-1/3*x^(-3)+1/5x^(-5)+C
5. ∫(lnx)^3/x^2dx
=-∫(lnx)^3(1/x)'dx
=-(lnx)^3(1/x)+3∫(lnx)^2(1/x)^2dx
=-(lnx)^3(1/x)-3∫(lnx)^2(1/x)'dx
=-(lnx)^3/x-3(lnx)^2/x+6∫lnx(1/x)^2dx
=-(lnx)^3/x-3(lnx)^2/x-6lnx/x+∫1/x^2dx
=-[(lnx)^3+3(lnx)^2+6lnx+1](1/x)+C.
那个。第三题能做吗
原式=∫x/(1+cosX)dx+∫sinX/(1+cosX)dx
=∫xsec^2(x/2)d(x/2)-∫1/(1+cosx)d(1+cosx)
=∫xd[tan(x/2)]-ln(1+cosx)
=xtan(x/2)-∫tan(x/2)dx-ln(1+cosx)
=xtan(x/2)+2ln∣cos(x/2)∣-ln2-ln∣cos(x/2)∣+C1
=xtan(x/2)+C