`int(1)/(sqrt(sec^(2)x+tan^(2)x))dx=`

Options

A. `-tan^(-1)(cosx)+c`
B. `log(cosx+sqrt(1+cos^(2)x))+c`
C. `log(sinx-sqrt(1+sin^(2)x))+c`
D. `log(sinx+sqrt(1+sin^(2)x))+c`

Answer

To solve the integral \( \int \frac{1}{\sqrt{\sec^2 x + \tan^2 x}} \, dx \), we can follow these steps:

### Step 1: Rewrite the expression inside the square root
We know that:
\[ \sec^2 x = 1 + \tan^2 x \]
Thus, we can rewrite the integral as:
\[ \int \frac{1}{\sqrt{\sec^2 x + \tan^2 x}} \, dx = \int \frac{1}{\sqrt{(1 + \tan^2 x) + \tan^2 x}} \, dx = \int \frac{1}{\sqrt{1 + 2\tan^2 x}} \, dx \]

### Step 2: Use trigonometric identities
We can express \(\tan^2 x\) in terms of \(\sin x\) and \(\cos x\):
\[ \tan^2 x = \frac{\sin^2 x}{\cos^2 x} \]
Substituting this back into the integral gives:
\[ \int \frac{1}{\sqrt{1 + 2\frac{\sin^2 x}{\cos^2 x}}} \, dx = \int \frac{1}{\sqrt{\frac{\cos^2 x + 2\sin^2 x}{\cos^2 x}}} \, dx \]
This simplifies to:
\[ \int \frac{\cos x}{\sqrt{\cos^2 x + 2\sin^2 x}} \, dx \]

### Step 3: Substitute \( t = \sin x \)
Let \( t = \sin x \), then \( dt = \cos x \, dx \). The integral becomes:
\[ \int \frac{1}{\sqrt{1 + 2t^2}} \, dt \]

### Step 4: Solve the integral
The integral \( \int \frac{1}{\sqrt{1 + 2t^2}} \, dt \) is a standard form. The result is:
\[ \frac{1}{\sqrt{2}} \ln \left| t + \sqrt{1 + 2t^2} \right| + C \]

### Step 5: Substitute back for \( t \)
Substituting back \( t = \sin x \):
\[ \frac{1}{\sqrt{2}} \ln \left| \sin x + \sqrt{1 + 2\sin^2 x} \right| + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{1}{\sqrt{\sec^2 x + \tan^2 x}} \, dx = \frac{1}{\sqrt{2}} \ln \left| \sin x + \sqrt{1 + 2\sin^2 x} \right| + C \]