How do you find the antiderivative of #3x^2 + sin(4x)+tan x sec x#?

1 Answer
Mar 11, 2015

The idea is to use a substitution. The notation varies, depending on what notation you've learned up to this point.

Be cause you are saying "antiderivative", rather than "integral" , I will assume that you haven't learned integral notation yet.

Part 1 Not too tough
So, the antiderivative of #x# to a power is the opposite of the derivative of #x# to a power. For the derivative, you multiply by the exponent, then make a new exponent by subtracting #1#. The opposite, then if first make a new exponent by adding #1# and then dividing (opposite of multiplying) by the new exponent.

So the antiderivative of #x^2# is #(x^3)/3#.

When we find derivatives any constant we're multiplying by just stays out front. It's the same for the antiderivative. So the antiderivative of #3x^2# is #3(x^3)/3#. Which simplifies to just #x^3#. (Check by differentiating #x^3#, Yes, we do get #3x^2#.

Part 2 Also not too tough
The antiderivative of #tanxsecx# is a function whose derivative is equal to #tanxsecx#. Think through the derivatives of trig functions you've memorized, #d/(dx)(sinx)=cosx# and so on. When you get to #d/(dx)(secx)=secxtanx# Stop!

#secxtanx=tanxsecx# so the antiderivative of #tanxsecx# is# secx#

Part 3 A little tougher, but you'll get used to it
(Really not bad, when you learn more notation.)
Now, for the middle, we need a function whose derivative is #sin(4x)#
There are a few ways of describing this next bit, here's one of them:

I've just thought about derivatives of trig function and one of them is #d/(dx)cosx=-sinx#. That's not exactly what I need, but it's close.

Why is it not right? Two reasons. We don't want the minus sign and we want #4x#, not just plain #x#.

So, what's the derivative of #cos(4x)#? It's#( -sin(4x))*4#

That's pretty close. In fact if we get rid of the minus sign and the extra #4#, we'll have it. Remember about the constant just staying out front when we differentiate?
Well, that mean is I use #1/4cos(4x)# the #4#'s will cancel after i find the derivative. To get rid of the minus, we'll stick a minus out front as well.

#d/(dx)(-1/4cos(4x))=sin4x#

Put the pieces together and add + C

The antiderivative of #3x^2+sin(4x)+tanxsecx# is #x^3-1/4cos(4x)+secx+C#

Wehn you learn other notation, you'll learn to call this "integration by substitution" -- though it will look a little different.