線上遊戲排行榜2013/2014,進擊的巨人線上看,candy crush saga外掛,正妹寫真三圍

1.∫sec^2x tan^3x dx

=∫tan^3(x)*d(tan(x));;d(tan(x))=sec^2(x)*dx

=(1/4)*tan^4(x)+C

2.∫sin^2x cos^2x dx

=(1/4)*∫[2sin(x)*cos(x)]^2*dx

=(1/4)*∫sin^2(2x)dx

=(1/8)*∫(1-cos(4x))dx

=(1/8){x-(1/4)*sin(4x)}+C

3.∫sin^5x dx

=-∫sin^4(x)d(cos(x))

=-∫(1-cos^2(x))^2*d(cos(x))

=-∫(1-2(cosx)^2+(cosx)^4)*d(cosx)

=∫(-1+2(cx)^2+(cx)^4)d(cx);;cx=cosx

=-cos(x)+(2/3)*cos^3(x)+(1/5)*cos^5(x)+C

4.∫dx/(9-x^2)^(5/2)

=∫3cos(y)dy/(3cos(y))^5;;y=3sin(x), dy=3cos(y)

=∫dy/(3cos(y))^4

=(1/81)*∫sec^4(y)dy

=(1/81)*∫sec^2(y)d(tan(y))

=(1/81)*∫(1+tan^2(y))d(tan(y))

=(1/81)*{tan(y)+(1/3)*tan^3(y)}+C

=(tan(y)/81)*(1+(1/3)*tan^2(y))+C

5.∫(4x^2-2x-10)dx/[(x+1)(x-1)x]

=∫[a/x+b/(x+1)+c/(x-1)]dx

=∫(10/x-2/(x+1)-4/(x-1))dx

=10*ln(x)-2*ln(x+1)-4*ln(x-1)+ln(c)

=ln{c*x^10/[(x+1)^2*(x-1)^4]}

=ln{cx^10/[(x-1)(x^2-1)]^2}

其中f(x)=2(2x^2-x-5)

=a(x^2-1)+bx(x-1)+cx(x+1)

f(0)=-10=-a => a=10

f(1)=-8=2c => c=-4

f(-1)=-4=2b => b=-2

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