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

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Answer 1 · 2016-01-07T01:50:05

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

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 $du = 2x$

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

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

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