∫cos3(7 x) dx

∫xarcsinxdx=∫arcsinxd(x²/2)=(1/2)x²arcsinx-(1/2)∫x²/√(1-x²)dx,x=sinz=(1/2)x²

∵∫e^(-x)cosxdx=e^(-x)sinx+∫e^(-x)sinxdx(应用分部积分法)==>∫e^(-x)cosxdx=e^(-x)sinx-e^(-x)cosx-∫e^(-x)cosxdx

解析tan'x=sec²x所以∫sec²xdx=tanx+c再问：∫sec^2tan^2dx等于多少呢再答：因为sec²xtan²x=sin²x∫si

分部积分法∫xcosxdx=∫xdsinx=xsinx-∫sinxdx=xsinx+cosx+C

∫xf''(x)dx=∫xdf'(x)=xf'(x)-∫f'(x)dx=xf'(x)-f(x)+C

=∫x(secx)^2dx=∫xdtanx=xtanx-∫tanxdx=xtanx-∫sinx/cosxdx=xtanx+∫dcosx/cosx=xtanx+ln|cosx|+C

∫e^(-x)dx=-∫e^(-x)d(-x)=-e^(-x)+c

原式=∫cos(3x)/sin（3x）dx=1/3∫1/sin(3x)dsin3x=1/3ln绝对值sin3x+c

∫(x²+7x+12)/(x+3)dx=∫(x+3)(x+4)/(x+3)dx=∫(x+4)dx=x平方/2+4

∫sin(5x)sin(7x)dx=(1/2)∫[cos(5x-7x)+cos(5x+7x)]dx=(1/2)∫[cos(2x)+cos(12x)]dx=(1/2)[(1/2)sin2x+(1/12)

∫dx/(e^x-e^(-x))=∫e^xdx/(e^2x-1)=∫1/(e^2x-1)de^x=1/2∫[1/(e^x-1)-1/(e^x+1)]de^x=1/2ln(e^x-1)-1/2ln(e^

∫1/(x^7+4x)dx=1/4*∫(x^6+4-x^6)/(x^7+4x)dx=1/4*[∫1/xdx-∫x^6/(x^7+4x)dx]=1/4*lnx-1/4*∫x^5/(x^6+4)dx=1/

2lnlx-4l-lnlx-3l+c∫(x-2)/(x^2-7x+12)dx=∫(x-2)/(x-4)(x-3)dx=∫[2/(x-4)-1/(x-3)]dx=∫2/(x-4)dx-∫1/(x-3)d

原式=1/2∫(sin(17x)+sin(-3x))dx=1/2∫sin(17x)dx-1/2∫sin(3x)dx=-1/34cos(17x)+1/6cos(6x)+C

若F(x)=∫f(x)dx则dF(x)=f(x)dx所以d(∫[sin(7x)]^5dx)={[sin(7x)]^5}dx

1、∫xsin(x^2)cos3(x^2)dx＝（1/2)∫sin(x^2)cos3(x^2)dx^2=(1/4)∫[sin4(x^2)-sin2(x^2)]dx^2=(1/4)[∫sin4(x^2)

1.∫(tanx+x)dx＝∫tanxdx+∫xdx2.∫tanxdx,令u=cosx,du=－sinxdx．∫tanxdx=-ln|cosx|+C．3.∫xdx＝x^2/2+c4.∫(tanx+x)

1.已知向量a=（sin3分之x,cos3分之x),b=(cos3分之x,根号3cos3分之x),函数f(x)=向量a·向量b则有：f(x)=sin(3分之x)cos(3分之x)+cos(3分之x)*

∫|7x-27|dx=∫（0,27/7)(-7x+27)dx+∫(27/7,10)(7x-27)dx=(-7/2X^2+27X)（0,27/7)+(7/2X^2-27X)（27/7,10)=184又1