How do I find the anti-derivative of x^3sec^2(x) + 3x^2tan(x)?

2 Answers
Apr 20, 2017

There's a trick to that one!

Explanation:

Let f(x) = tanx and g(x) = x^3.
Now observe that
x^3sec^2x+3x^2tanx
is equal to
f'(x)g(x) + f(x)g'(x).
By the Product Rule for differentiation, it is the derivative of
f(x)g(x) = x^3tanx.
Don't forget the +C if the integral is indefinite!

Apr 20, 2017

x^3tan(x)+C

Explanation:

We have:

int(x^3sec^2(x)+3x^2tan(x))dx

=intx^3sec^2(x)dx+int3x^2tan(x)dx

Just focusing on the second integral, we should set this up for integration by parts. Let:

{(u=tan(x),=>,du=sec^2(x)dx),(dv=3x^2dx,=>,v=x^3):}

Then:

=intx^3sec^2(x)dx+uv-intvdu

=intx^3sec^2(x)dx+x^3tan(x)-intx^3sec^2(x)dx

=x^3tan(x)+C