The value of `int(secxdx)/(sqrt(sin(2x+theta)+sintheta))` is

Options

A. `sqrt((tanx+tantheta)sec theta)+C`
B. `sqrt(2(tanx+tantheta)sectheta)+C`
C. `sqrt(2(tanx+tan)sectheta)+C`
D. none of these

Answer

To solve the integral

\[ I = \int \frac{\sec x \, dx}{\sqrt{\sin(2x + \theta) + \sin \theta}} \]

we will follow a series of steps to simplify and evaluate the integral.

### Step 1: Rewrite the Integral
Using the identity for sine of a sum, we can express \(\sin(2x + \theta)\) as:

\[ \sin(2x + \theta) = \sin 2x \cos \theta + \cos 2x \sin \theta \]

Thus, we can rewrite the integral as:

\[ I = \int \frac{\sec x \, dx}{\sqrt{\sin 2x \cos \theta + \cos 2x \sin \theta + \sin \theta}} \]

### Step 2: Simplify the Denominator
We can factor out \(\sin \theta\) from the terms in the square root:

\[ I = \int \frac{\sec x \, dx}{\sqrt{\sin 2x \cos \theta + \sin \theta (\cos 2x + 1)}} \]

### Step 3: Use Trigonometric Identities
Recall that \(\cos 2x = 2\cos^2 x - 1\). Therefore, we can rewrite \(\cos 2x + 1\) as:

\[ \cos 2x + 1 = 2 \cos^2 x \]

Substituting this back into our integral gives:

\[ I = \int \frac{\sec x \, dx}{\sqrt{\sin 2x \cos \theta + 2 \sin \theta \cos^2 x}} \]

### Step 4: Substitute for \(\sin 2x\)
Using the identity \(\sin 2x = 2 \sin x \cos x\), we can rewrite the integral as:

\[ I = \int \frac{\sec x \, dx}{\sqrt{2 \sin x \cos x \cos \theta + 2 \sin \theta \cos^2 x}} \]

### Step 5: Factor Out Common Terms
We can factor out \(2\) from the square root:

\[ I = \int \frac{\sec x \, dx}{\sqrt{2} \sqrt{\sin x \cos x \cos \theta + \sin \theta \cos^2 x}} \]

### Step 6: Simplify Further
Now, we can express \(\sec x\) as \(\frac{1}{\cos x}\):

\[ I = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{\sin x \cos x \cos \theta + \sin \theta \cos^2 x}} \]

### Step 7: Change of Variables
Let \(t = \tan x\), then \(dx = \frac{dt}{\sec^2 x} = \frac{dt}{1 + t^2}\). The integral becomes:

\[ I = \frac{1}{\sqrt{2}} \int \frac{dt}{\sqrt{\frac{t}{1+t^2} \cos \theta + \sin \theta \frac{1}{1+t^2}}} \]

### Step 8: Evaluate the Integral
This integral can be evaluated using standard techniques, leading to:

\[ I = \frac{1}{\sqrt{2} \cos \theta} \left( \sqrt{t + \tan \theta} \right) + C \]

### Final Result
Thus, the value of the integral is:

\[ I = \frac{1}{\sqrt{2} \cos \theta} \left( \tan x + \tan \theta \right) + C \]