`int(costheta)/(7+sintheta-2cos^2theta)d theta` is equal to

Options

A. `2/sqrt(39)tan^(-1)((2sintheta+1)/sqrt(39))+C`
B. `2/sqrt(39)tan^(-1)((4sintheta+1)/sqrt(39))+C`
C. `4/sqrt(39)tan^(-1)((4sintheta+1)/sqrt(39))+C`
D. `4/sqrt(39)tan^(-1)((2sintheta+1)/sqrt(39))+C`

Answer

To solve the integral

\[ I = \int \frac{\cos \theta}{7 + \sin \theta - 2 \cos^2 \theta} \, d\theta, \]

we will follow these steps:

### Step 1: Rewrite the Denominator
First, we can rewrite \(2 \cos^2 \theta\) using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\):

\[ 2 \cos^2 \theta = 2(1 - \sin^2 \theta) = 2 - 2 \sin^2 \theta. \]

Thus, the denominator becomes:

\[ 7 + \sin \theta - 2 \cos^2 \theta = 7 + \sin \theta - (2 - 2 \sin^2 \theta) = 5 + \sin \theta + 2 \sin^2 \theta. \]

### Step 2: Substitute \( \sin \theta = t \)
Let \( t = \sin \theta \). Then, \( d\theta = \frac{dt}{\cos \theta} \). Since \(\cos^2 \theta = 1 - \sin^2 \theta = 1 - t^2\), we have \(\cos \theta = \sqrt{1 - t^2}\). Therefore, the integral becomes:

\[ I = \int \frac{\sqrt{1 - t^2}}{5 + t + 2t^2} \cdot \frac{dt}{\sqrt{1 - t^2}} = \int \frac{1}{5 + t + 2t^2} \, dt. \]

### Step 3: Completing the Square
Next, we need to complete the square in the denominator \(5 + t + 2t^2\). We can rewrite it as:

\[ 2t^2 + t + 5 = 2\left(t^2 + \frac{1}{2}t + \frac{5}{2}\right). \]

To complete the square inside the parentheses:

\[ t^2 + \frac{1}{2}t = \left(t + \frac{1}{4}\right)^2 - \frac{1}{16}. \]

Thus,

\[ 5 + t + 2t^2 = 2\left(\left(t + \frac{1}{4}\right)^2 + \frac{39}{16}\right). \]

### Step 4: Substitute Back into the Integral
Now, substituting this back into the integral, we have:

\[ I = \frac{1}{2} \int \frac{1}{\left(t + \frac{1}{4}\right)^2 + \left(\frac{\sqrt{39}}{4}\right)^2} \, dt. \]

### Step 5: Use the Arctangent Formula
The integral of the form \(\int \frac{1}{x^2 + a^2} \, dx\) is \(\frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + C\). Here, \(x = t + \frac{1}{4}\) and \(a = \frac{\sqrt{39}}{4}\):

\[ I = \frac{1}{2} \cdot \frac{4}{\sqrt{39}} \tan^{-1} \left(\frac{4(t + \frac{1}{4})}{\sqrt{39}}\right) + C. \]

### Step 6: Substitute Back \(t = \sin \theta\)
Substituting back \(t = \sin \theta\):

\[ I = \frac{2}{\sqrt{39}} \tan^{-1} \left(\frac{4 \sin \theta + 1}{\sqrt{39}}\right) + C. \]

### Final Answer
Thus, the final result of the integral is:

\[ I = \frac{2}{\sqrt{39}} \tan^{-1} \left(\frac{4 \sin \theta + 1}{\sqrt{39}}\right) + C. \]