Question #c5bb3

3 Answers
Dec 20, 2017

3) int (arctanx)^2/(1+x^2)*dx

=(arctanx)^3/3+C

Explanation:

3) int (arctanx)^2/(1+x^2)*dx

After using u=arctanx, x=tanu and dx=(secu)^2*du transforms, this integral became,

int (u^2*(secu)^2*du)/(secu)^2

=u^2*du

=u^3/3+C

=(arctanx)^3/3+C

Dec 20, 2017

1) int (x^2-3x)*(cosx)^2*dx

=x^3/6-(3x^2)/4+(2x^2-6x-1)/8*sin2x+(2x-3)/8*cos2x+C

Explanation:

1) int (x^2-3x)*(cosx)^2*dx

=1/2int (x^2-3x)*(1+cos2x)*dx

=1/2int (x^2-3x)*dx+1/2int (x^2-3x)*cos2x*dx

A=int (x^2-3x)*dx=x^3/3-(3x^2)/2+2C

B=int (x^2-3x)*cos2x

=(x^2-3x)*1/2sin2x-(2x-3)(-1/4cos2x)+2(-1/8sin2x)

=(2x^2-6x-1)/4*sin2x+(2x-3)/4*cos2x

Thus,

int (x^2-3x)*(cosx)^2*dx

=1/2*A+1/2*B

=x^3/6-(3x^2)/4+(2x^2-6x-1)/8*sin2x+(2x-3)/8*cos2x+C

Dec 20, 2017

int (x^3+2)/(x^2-x+1)*dx

=x^2/2+x+(2sqrt3)/3arctan((2x-1)/sqrt3)+C

Explanation:

2) int (x^3+2)/(x^2-x+1)*dx

=int (x^3+1)/(x^2-x+1)*dx+int (dx)/(x^2-x+1)

=int ((x^2-x+1)*(x+1))/(x^2-x+1)*dx+int (4dx)/(4x^2-4x+4)

=int (x+1)*dx+2int (2dx)/((2x-1)^2+3)

=x^2/2+x+(2sqrt3)/3arctan((2x-1)/sqrt3)+C