How do you integrate #int x^3 t an x dx # using integration by parts?

1 Answer
Jan 7, 2016

#intx^3tan(x)dx = ln|cos(x)|(x^3 - 3x^2 + 6x - 6 + c)#

Explanation:

Say #dv = tan(x)# so #v = ln|cos(x)|#
And #u = x^3# so #du = 3x^2#

#intx^3tan(x)dx = x^3*ln|cos(x)| - 3inttan(x)*x^2dx#

For the latter integral, repeat

#dv = tan(x)# so #v = ln|cos(x)|#, #u = x^2# so #du = 2x#

#intx^3tan(x)dx = x^3*ln|cos(x)| - 3(x^2ln|cos(x)| - 2inttan(x)xdx)#
#intx^3tan(x)dx = x^3*ln|cos(x)| - 3x^2ln|cos(x)| + 6inttan(x)xdx#

And once more to finish it

#dv = tan(x)# so #v = ln|cos(x)|#, #u = x# so #du = 1#

#intx^3tan(x)dx = x^3ln|cos(x)| - 3x^2ln|cos(x)| + 6(xln|cos(x)| - ln|cos(x)| + c)#

#intx^3tan(x)dx = x^3ln|cos(x)| - 3x^2ln|cos(x)| + 6xln|cos(x)| - 6ln|cos(x)| + c#

Or

#intx^3tan(x)dx = ln|cos(x)|(x^3 - 3x^2 + 6x - 6 + c)#