How do you integrate int x^3 t an x dx ∫x3tanxdx using integration by parts?

1 Answer
Lovecraft
Jan 7, 2016

intx^3tan(x)dx = ln|cos(x)|(x^3 - 3x^2 + 6x - 6 + c)∫x3tan(x)dx=ln|cos(x)|(x3−3x2+6x−6+c)

Explanation:

Say dv = tan(x)dv=tan(x) so v = ln|cos(x)|v=ln|cos(x)|
And u = x^3u=x3 so du = 3x^2du=3x2

intx^3tan(x)dx = x^3*ln|cos(x)| - 3inttan(x)*x^2dx∫x3tan(x)dx=x3⋅ln|cos(x)|−3∫tan(x)⋅x2dx

For the latter integral, repeat

dv = tan(x)dv=tan(x) so v = ln|cos(x)|v=ln|cos(x)|, u = x^2u=x2 so du = 2xdu=2x

intx^3tan(x)dx = x^3*ln|cos(x)| - 3(x^2ln|cos(x)| - 2inttan(x)xdx)∫x3tan(x)dx=x3⋅ln|cos(x)|−3(x2ln|cos(x)|−2∫tan(x)xdx)
intx^3tan(x)dx = x^3*ln|cos(x)| - 3x^2ln|cos(x)| + 6inttan(x)xdx∫x3tan(x)dx=x3⋅ln|cos(x)|−3x2ln|cos(x)|+6∫tan(x)xdx

And once more to finish it

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

intx^3tan(x)dx = x^3ln|cos(x)| - 3x^2ln|cos(x)| + 6(xln|cos(x)| - ln|cos(x)| + c)∫x3tan(x)dx=x3ln|cos(x)|−3x2ln|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∫x3tan(x)dx=x3ln|cos(x)|−3x2ln|cos(x)|+6xln|cos(x)|−6ln|cos(x)|+c

Or

intx^3tan(x)dx = ln|cos(x)|(x^3 - 3x^2 + 6x - 6 + c)∫x3tan(x)dx=ln|cos(x)|(x3−3x2+6x−6+c)

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