What is `int (a + b sin x)/(cos^(2) x) dx` equal to ?

Options

A. `a sec x + b tan x + c`
B. `a tan x + b sec x + c`
C. `a cot x + b "cosec" x + c`
D. `a "cosec" x + b cot x + c`

Explanation

To solve the integral \(\int \frac{a + b \sin x}{\cos^2 x} \, dx\), we can break it down into simpler parts. Here’s the step-by-step solution:

### Step 1: Split the Integral
We can separate the integral into two parts:
\[ \int \frac{a + b \sin x}{\cos^2 x} \, dx = \int \frac{a}{\cos^2 x} \, dx + \int \frac{b \sin x}{\cos^2 x} \, dx \]

### Step 2: Rewrite Using Trigonometric Identities
We know that \(\frac{1}{\cos^2 x} = \sec^2 x\) and \(\frac{\sin x}{\cos^2 x} = \sec^2 x \tan x\). Thus, we can rewrite the integrals:
\[ \int \frac{a}{\cos^2 x} \, dx = a \int \sec^2 x \, dx \]
\[ \int \frac{b \sin x}{\cos^2 x} \, dx = b \int \sec^2 x \tan x \, dx \]

### Step 3: Integrate Each Part
Now we can integrate each part:
1. The integral of \(\sec^2 x\) is \(\tan x\):
\[ a \int \sec^2 x \, dx = a \tan x \]
2. The integral of \(\sec^2 x \tan x\) is \(\sec x\):
\[ b \int \sec^2 x \tan x \, dx = b \sec x \]

### Step 4: Combine the Results
Combining both results, we have:
\[ \int \frac{a + b \sin x}{\cos^2 x} \, dx = a \tan x + b \sec x + C \]
where \(C\) is the constant of integration.

### Final Answer
Thus, the final result is:
\[ \int \frac{a + b \sin x}{\cos^2 x} \, dx = a \tan x + b \sec x + C \]

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