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

Options

A. `asecx+btanx+c`
B. `atanx+bsecx+c`
C. `acotx+bcosecx+c`
D. `acosecx+bcotx+c`

Answer

To solve the integral \(\int \frac{a + b \sin x}{\cos^2 x} \, dx\), we can break it down into two separate integrals. Here are the steps:

### Step 1: Separate the Integral
We can rewrite the integral as:
\[ \int \frac{a + b \sin x}{\cos^2 x} \, dx = \int \frac{a}{\cos^2 x} \, dx + \int \frac{b \sin x}{\cos^2 x} \, dx \]

### Step 2: Simplify Each Integral
The first integral can be simplified:
\[ \int \frac{a}{\cos^2 x} \, dx = a \int \sec^2 x \, dx \]
The second integral can be simplified using the identity \(\tan x = \frac{\sin x}{\cos x}\):
\[ \int \frac{b \sin x}{\cos^2 x} \, dx = b \int \tan x \sec x \, dx \]

### 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 \(\tan x \sec x\) is \(\sec x\):
\[ b \int \tan x \sec x \, dx = b \sec x \]

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

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