How do I evaluate the indefinite integral intx*sin(x)*tan(x)dx ?

1 Answer
Oct 24, 2014

This integral can be evaluated using integration by parts .

Firstly, integrating by parts where u=x and dv=sin(x)dx, we get
int xsin(x)dx=-x cos(x)+sin(x)+c

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

int xcos (x)dx=x sin (x)-sin (x)+c

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

int x sin(x) tan(x)dx=
=-x tan(x)cos(x)+sin(x)tan(x)+int xcos(x)sec^2(x)dx-int sin(x)sec^2(x)dx
=-xsin(x)+sin(x)tan(x)+int xcos(x)dx-int sin(x)/cos^2(x)dx
=-xsin(x)+sin(x)tan(x)+x sin (x)-sin (x)-int sin(x)/cos^2(x)dx
=sin(x)tan(x)-sin (x)-int sin(x)/(cos^2(x))dx
Finally, using substitution u=cos(x), du=-sin(x) dx in the last integral we have

int x sin(x) tan(x)dx=sin(x)tan(x)-sin (x)+int 1/u^2du
=sin(x)tan(x)-sin (x)-1/3cos^3(x)+c.