Why does the antiderivative of $\int (1+u^2)u$ become $(u+ \frac{1}{3}u^3) + C$?

Question

Example problem from Calc II lecture: $\int \sec^4(x)dx$

Worked out solution from lecture:

$\int \sec^4(x)dx = \int (1+\tan^2(x))(\sec^2(x))dx = \int (1+u^2)du = $

Question: Why when doing the antiderivative does the 1 in $\int (1+u^2)du$ become $u$ rather than $x$.

Based on my current working knowledge of Calculus, the antiderivative of 1 is always $x$. Does substitution have anything to do with this?

Thanks

Answer 1

Look closely. It isn’t $\int (1+u^2)dx$, it’s $\int (1+u^2)du$. The antiderivative of $du$ is of course $u$, not $x$. Does that answer your question?

Answer 2

To answer your question, consider what exactly is meant by the antiderivative. By definition, it is that $\frac {d}{dx}\int sec^4(x)dx=sec^4x$

By chain rule, we get

$\frac{d}{dx}tan(x)\times \frac{d}{d tan(x)}\int sec^4(x)dx=sec^4(x)=(1+tan^2x)sec^2x$

Dividing $\frac{d}{dx}tan(x)=sec^2x$ from both sides, we get

$\frac{d}{dtan(x)}\int sec^4dx=1+tan^2x$

Let $tan(x)=u$.

$\frac{d}{du}\int sec^4x dx=1+u^2$

Now notice that this form is precisely what is used to define an integral! So we get that we just need to find $\int (1+u^2)du.$

The value of this antiderivative is $u+\frac{u^3}{3}$. If you're wondering why it is a $u$ and not $x$, that is because this time it is $du$ not $dx$.

Answer 3

Let

$$I=\int\sec^4xdx.$$

By rewriting

$$\sec^4x=(1+\tan^2x)\sec^2x.$$

We set $u=\tan x, du = \sec^2 x dx$.

Thus $I$ becomes

\begin{align} I&=\int \sec^4x dx\\ &=\int \sec^2x\sec^2xdx && \text{properties of exponents}\\ &=\int (1+\tan^2x)\sec^2xdx && \text{trig identity}\\ &=\int (1+u^2)\sec^2x\frac{1}{\sec^2x}du&& \text{$u-$substitution}\\ &=\int 1+u^2 du && \text{cancellation due to $u-$sub}\\ &=u+\frac{1}{3}u^3+C && \text{power rule for integration}\\ &=\tan x + \frac{1}{3}\tan x +C && \text{swapping back in $x$'s for $u$'s}. \end{align}

Here, $C$ is your constant of integration.