Q.Integrate the following function: $x \sec^2 x$

The key idea is integration by parts, treating $x$ as the first function and $\sec^2 x$ as the second function.

Let $u = x$ and $dv = \sec^2 x \, dx$. Then $du = dx$ and $v = \tan x$.

Using the formula $\int u \, dv = uv - \int v \, du$:

$\int x \sec^2 x \, dx = x \tan x - \int \tan x \, dx$ …

Why integration by parts?

When you see a product of two different kinds of functions — here, a polynomial ($x$) and a trig function ($\sec^2 x$) — the Power Rule alone won't help. You need a technique that reverses the product rule for derivatives. That's exactly what integration by parts does.

The formula is:

The art is in choosing $u$ and $dv$ wisely. A good rule of thumb: pick $u$ as the function that simplifies when differentiated, and $dv$ as the part you can integrate easily.

Here, $x$ becomes simpler when differentiated (it becomes $1$), and $\sec^2 x$ is straightforward to integrate (it's $\tan x$). Perfect match.

Step-by-step solution

1. Set up the parts

Let:

Differentiate $u$ and integrate $dv$:

2. Apply the formula

Substitute into $\int u \, dv = uv - \int v \, du$:

3. Handle the remaining integral

Now we need $\int \tan x \, dx$. This is a standard result, but let's derive it quickly:

Let $t = \cos x$, then $dt = -\sin x \, dx$, so:

$\int \frac{\sin x}{\cos x} \, dx = -\int \frac{1}{t} \, dt = -\log|t| + C = -\log|\cos x| + C$ …

Method: Integration by parts, algebraic times trigonometric

For $x\cdot(\text{trig})$ such as $x\sec^{2}x$, take the algebraic factor as $u$ so that differentiating it lowers the power, and integrate the trig factor.

Steps

Step 1: Assign parts (ILATE: Algebraic before Trigonometric).

Differentiating $x$ gives the simple $1$; $\sec^{2}x$ integrates cleanly.

Step 2: Compute $du$ and $v$.

Step 3: Apply the formula. …

Common Mistakes

Mistake 1: Taking $u=\sec^{2}x$, $dv=x\,dx$.

Why it's wrong: differentiating $\sec^{2}x$ makes the integrand messier, not simpler. Correct approach: let the algebraic $x$ be $u$.

Mistake 2: Writing $\int\tan x\,dx=\log|\tan x|$ or $\sec^{2}x$.

Why it's wrong: $\int\tan x\,dx=-\log|\cos x|+C=\log|\sec x|+C$. Correct approach: recall the standard result and its sign. …