Integral Value Calculation: Solving \( \int \frac{x}{\cos^2 x} dx \)

To find the value of the integral \( \int \frac{x}{\cos^2 x} dx \), we can rewrite the expression and use the technique of integration by parts.

First, we can rewrite \( \frac{1}{\cos^2 x} \) as \( \sec^2 x \). So, the integral becomes:

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

Integration by Parts Explained

This integral is in the form suitable for integration by parts. The integration by parts formula is given by:

\[ \int u \, dv = uv - \int v \, du \]

We need to choose appropriate functions for \(u\) and \(dv\). A common mnemonic for choosing \(u\) is LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential). In this case, \(x\) is algebraic and \( \sec^2 x \) is trigonometric.

• Let \(u = x\) (Algebraic function)
• Let \(dv = \sec^2 x \, dx\) (Trigonometric function)

Step-by-Step Derivation

Now, we need to find \(du\) and \(v\) from our chosen \(u\) and \(dv\):

• Differentiate \(u\) to find \(du\):

  \[ du = \frac{d}{dx}(x) \, dx = dx \]

• Integrate \(dv\) to find \(v\):

  \[ v = \int \sec^2 x \, dx = \tan x \]

Now, substitute these into the integration by parts formula \( \int u \, dv = uv - \int v \, du \):

\[ \int x \sec^2 x \, dx = x (\tan x) - \int (\tan x) (dx) \]

This simplifies to:

\[ \int x \sec^2 x \, dx = x \tan x - \int \tan x \, dx \]

Evaluating the Remaining Integral

Next, we need to evaluate the integral \( \int \tan x \, dx \). We know that \( \tan x = \frac{\sin x}{\cos x} \). So, the integral becomes:

\[ \int \frac{\sin x}{\cos x} \, dx \]

To solve this, we can use a simple substitution. Let \(t = \cos x\). Then, differentiate \(t\) with respect to \(x\):

\[ \frac{dt}{dx} = -\sin x \implies dt = -\sin x \, dx \]

So, \( \sin x \, dx = -dt \). Substitute these into the integral:

\[ \int \frac{-dt}{t} = -\int \frac{1}{t} \, dt \]

Integrating \( \frac{1}{t} \) with respect to \(t\) gives \( \log|t| \). So:

\[ -\int \frac{1}{t} \, dt = -\log|t| + C \]

Now, substitute back \(t = \cos x\):

\[ -\log|\cos x| + C \]

We can also write \( -\log|\cos x| \) as \( \log\left(\frac{1}{|\cos x|}\right) \) or \( \log|\sec x| \).

Combining the Results

Substitute the result of \( \int \tan x \, dx \) back into our main integration by parts expression:

\[ \int x \sec^2 x \, dx = x \tan x - (-\log|\cos x|) + C \]

\[ \int x \sec^2 x \, dx = x \tan x + \log|\cos x| + C \]

When dealing with trigonometric functions in indefinite integrals like this, often the absolute value is omitted if the domain is assumed where the function is positive. Thus, the solution can be written as:

\[ x \tan x + \log (\cos x) + C \]

Comparing with Options

Let's compare our derived solution with the given options:

• Option 1: \(x \tan x\)
• Option 2: \(x \tan x + \log (\cos x)\)
• Option 3: \( \log \cos x \)
• Option 4: \(x \tan x – \log (\cos x)\)

Our calculated value \( x \tan x + \log (\cos x) + C \) matches Option 2 (ignoring the constant of integration, which is usually implied in such MCQ options).