How do you integrate int x^3 / ((sqrt(16+x^2))^3) dx∫x3(√16+x2)3dx using trigonometric substitution?
1 Answer
Explanation:
Remember that
For the case is convenient that
x=4tanyx=4tany
dx=4sec^2 y *dydx=4sec2y⋅dy
Then
int x^3/(sqrt(16+x^2))^3dx=int (64tan^3 y)/(sqrt(16+16tan^2 y))^3*4sec^2y* dy∫x3(√16+x2)3dx=∫64tan3y(√16+16tan2y)3⋅4sec2y⋅dy
=int (256 tan^3 y*sec^2 y)/(4sec y)^3dy=int (4 tan^3 y)/sec y dy=∫256tan3y⋅sec2y(4secy)3dy=∫4tan3ysecydy
=4*int sin^3 y/cos^3 y*cos y*dy=4int sin y/cos^2 y*(1-cos^2 y) dy=4⋅∫sin3ycos3y⋅cosy⋅dy=4∫sinycos2y⋅(1−cos2y)dy
=4int sin y/cos^2 y*dy-4int sin y*dy=4∫sinycos2y⋅dy−4∫siny⋅dy [A]
-> int sin y/cos^2 y*dy=→∫sinycos2y⋅dy=
Makingcos y=zcosy=z and-sin y*dy=dz−siny⋅dy=dz
=- int dz/z^2=1/z=1/cos y=−∫dzz2=1z=1cosy Inserting the result above in expression [A]:
=4/cos y+4cos y +const.=4cosy+4cosy+const. [B]
But
4tan y=x4tany=x =>siny=x/4*cos ysiny=x4⋅cosy
-> cos^2 y+sin^2 y = 1→cos2y+sin2y=1 =>cos^2 y+x^2/16*cos^2 y=1cos2y+x216⋅cos2y=1 =>(1+x^2/16)cos^2 y=1(1+x216)cos2y=1 =>cos y=sqrt(16/(16+x^2))=4/sqrt(16+x^2)cosy=√1616+x2=4√16+x2
Inserting the result above in expression [B]:
=cancel(4)/(cancel(4)/sqrt(16+x^2))+4*4/sqrt(16+x^2)+const.
=sqrt(16+x^2)+16/sqrt(16+x^2)+const.
