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

2 Answers
Max G.
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!

mason m
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#

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