How do I evaluate the indefinite integral xsin(x)tan(x)dx ?

1 Answer
Oct 24, 2014

This integral can be evaluated using integration by parts .

Firstly, integrating by parts where u=xanddv=sin(x)dx, we get
xsin(x)dx=xcos(x)+sin(x)+c

In the same way, for u=xanddv=cos(x)dx, we have

xcos(x)dx=xsin(x)sin(x)+c

Now, using integration by parts again with u=tan(x)anddv=xsin(x)dx and the first integral we have

xsin(x)tan(x)dx=
=xtan(x)cos(x)+sin(x)tan(x)+xcos(x)sec2(x)dxsin(x)sec2(x)dx
=xsin(x)+sin(x)tan(x)+xcos(x)dxsin(x)cos2(x)dx
=xsin(x)+sin(x)tan(x)+xsin(x)sin(x)sin(x)cos2(x)dx
=sin(x)tan(x)sin(x)sin(x)cos2(x)dx
Finally, using substitution u=cos(x),du=sin(x)dx in the last integral we have

xsin(x)tan(x)dx=sin(x)tan(x)sin(x)+1u2du
=sin(x)tan(x)sin(x)13cos3(x)+c.