What is the integral of sec^2(x)?

The Mathematician — Sat, 14 Jan 2023 13:00:10 +0000

The integral of

\sec^2(ax)

\frac{1}{a}\tan(ax) + C

.

Proof. In this case, it is easy to show what the integral

\sec^2(ax)

is if you have seen derivatives. Notice that we have seen here that

\begin{align*} \frac{d}{dx} \tan(x) = \sec^2(x). \end{align*}

Taking the antiderivative of

\sec^2(x)

, we do directly see that:

\begin{align*} \int \sec^2(x)dx = \tan(x) + C'. \end{align*}

What about the integral of

\sec^2(ax)

? We do know that the derivative of

\frac{1}{a}\tan(ax)

\sec^2(ax)

by using the chain rule. Therefore, taking antiderivative of

\sec^2(ax)

, we get:

\begin{align*} \int \sec^2(ax)dx = \frac{1}{a}\tan(ax) + C. \end{align*}

So, the integral of

\sec^2(ax)

\frac{1}{a}\tan(ax) + C

What is the Derivative of sec^2(x)?

The Mathematician — Mon, 31 Oct 2022 13:00:05 +0000

The derivative of

\sec^2(x)

2\tan(x)\sec^2(x)

.

Solution. Let

F(x) = \sec^2(x)

f(u) = u^2

and

g(x) = \sec(x)

. Then we will apply the chain rule:

\begin{align*} F'(x) = f'(g(x))g'(x). \end{align*}

We have seen here that

\frac{d}{dx} \sec(x) = \tan(x)\sec(x)

. So we get:

\begin{align*} f'(u) = 2u \quad \text{and} \quad g'(x) = \tan(x)\sec(x). \end{align*}

Combining everything, we get:

\begin{align*} F'(x) &= f'(g(x))g'(x) \\ &= 2\sec(x)\tan(x)\sec(x) \\ &= 2\tan(x)\sec^2(x). \end{align*}

Therefore, the derivative of

\sec^2(x)

2\tan(x)\sec^2(x)