Q. \(\displaystyle \int \sec^3 x\, dx\)

We want to find the indefinite integral

\[
\int \sec^3 x \, dx
\]

Step 1: Use the identity to rewrite \( \sec^3 x \).

We can factor out one \(\sec x\):

\[
\sec^3 x = \sec x \cdot \sec^2 x
\]

Step 2: Use the known derivative relationships involving \(\tan x\).

Recall:

• \(\displaystyle \frac{d}{dx}(\tan x) = \sec^2 x\)
• \(\displaystyle \sec^2 x = 1 + \tan^2 x\)

We will separate \(\sec^3 x\) into a form where \(\sec^2 x\,dx\) appears.

Write:

\[
\int \sec^3 x \, dx = \int \sec x \sec^2 x \, dx
\]

Step 3: Substitute using \(u = \tan x\).

Let

\[
u = \tan x
\]

Then

\[
\frac{du}{dx} = \sec^2 x
\quad \Rightarrow \quad
du = \sec^2 x \, dx
\]

Now rewrite the integral:

\[
\int \sec x \sec^2 x \, dx = \int \sec x \, du
\]

Step 4: Express \(\sec x\) in terms of \(u\).

We know the identity:

\[
\sec^2 x = 1 + \tan^2 x
\]

Since \(u = \tan x\), this becomes:

\[
\sec^2 x = 1 + u^2
\]

So

\[
\sec x = \sqrt{1+u^2}
\]

(This is the standard algebraic form; the final answer is valid up to the same sign conventions under the square root.)

Step 5: Integrate.

The integral becomes:

\[
\int \sec x \, du = \int \sqrt{1+u^2}\, du
\]

There is a standard result:

\[
\int \sqrt{1+u^2}\, du
=
\frac{1}{2}\left(u\sqrt{1+u^2} + \ln\left|u+\sqrt{1+u^2}\right|\right) + C
\]

Apply this:

\[
\int \sqrt{1+u^2}\, du
=
\frac{1}{2}\left(u\sqrt{1+u^2} + \ln\left|u+\sqrt{1+u^2}\right|\right) + C
\]

Step 6: Substitute back \(u = \tan x\).

We have:

• \(u = \tan x\)
• \(\sqrt{1+u^2} = \sqrt{1+\tan^2 x} = \sec x\)

So \(u\sqrt{1+u^2}\) becomes:

\[
u\sqrt{1+u^2} = \tan x \cdot \sec x
\]

And the logarithm becomes:

\[
\ln\left|u+\sqrt{1+u^2}\right|
=
\ln\left|\tan x + \sec x\right|
\]

Therefore the integral is:

\[
\int \sec^3 x \, dx
=
\frac{1}{2}\left(\sec x \tan x + \ln\left|\sec x + \tan x\right|\right) + C
\]

Final answer:

\[
\boxed{\int \sec^3 x \, dx = \frac{1}{2}\left(\sec x \tan x + \ln\left|\sec x + \tan x\right|\right) + C}
\]