Evaluate `int(dx)/(sinx(3+cos^(2)x))`

Explanation

To evaluate the integral \( I = \int \frac{dx}{\sin x (3 + \cos^2 x)} \), we will follow these steps:

### Step 1: Multiply and Divide by \(\sin x\)
We start by rewriting the integral:
\[ I = \int \frac{\sin x \, dx}{\sin^2 x (3 + \cos^2 x)} \]
This allows us to express \(\sin^2 x\) in terms of \(\cos^2 x\).

### Step 2: Use the Pythagorean Identity
Using the identity \(\sin^2 x = 1 - \cos^2 x\), we can rewrite the integral as:
\[ I = \int \frac{\sin x \, dx}{(1 - \cos^2 x)(3 + \cos^2 x)} \]

### Step 3: Substitution
Let \( t = \cos x \). Then, \( dt = -\sin x \, dx \) or \( \sin x \, dx = -dt \). The integral becomes:
\[ I = -\int \frac{dt}{(1 - t^2)(3 + t^2)} \]

### Step 4: Simplify the Integral
We can rewrite the integral as:
\[ I = -\int \frac{dt}{(3 + t^2)(1 - t^2)} \]

### Step 5: Partial Fraction Decomposition
We will decompose this into partial fractions:
\[ \frac{1}{(3 + t^2)(1 - t^2)} = \frac{A}{3 + t^2} + \frac{B}{1 - t^2} \]
Multiplying through by the denominator gives:
\[ 1 = A(1 - t^2) + B(3 + t^2) \]
Expanding and collecting like terms, we get:
\[ 1 = (B - A)t^2 + (3B + A) \]
From this, we can set up the equations:
1. \( B - A = 0 \)
2. \( 3B + A = 1 \)

### Step 6: Solve for A and B
From \( B - A = 0 \), we have \( B = A \). Substituting into the second equation:
\[ 3A + A = 1 \implies 4A = 1 \implies A = \frac{1}{4}, B = \frac{1}{4} \]

### Step 7: Substitute Back into the Integral
Now we can rewrite the integral:
\[ I = -\int \left( \frac{1/4}{3 + t^2} + \frac{1/4}{1 - t^2} \right) dt \]
This gives:
\[ I = -\frac{1}{4} \int \frac{dt}{3 + t^2} - \frac{1}{4} \int \frac{dt}{1 - t^2} \]

### Step 8: Evaluate Each Integral
1. The first integral:
\[ \int \frac{dt}{3 + t^2} = \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{t}{\sqrt{3}}\right) \]
2. The second integral:
\[ \int \frac{dt}{1 - t^2} = \frac{1}{2} \ln\left|\frac{1+t}{1-t}\right| \]

### Step 9: Combine Results
Putting it all together, we have:
\[ I = -\frac{1}{4} \left( \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{t}{\sqrt{3}}\right) + \frac{1}{2} \ln\left|\frac{1+t}{1-t}\right| \right) + C \]

### Step 10: Substitute Back for \(t\)
Substituting back \( t = \cos x \):
\[ I = -\frac{1}{4} \left( \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{\cos x}{\sqrt{3}}\right) + \frac{1}{2} \ln\left|\frac{1+\cos x}{1-\cos x}\right| \right) + C \]

### Final Answer
Thus, the evaluated integral is:
\[ I = -\frac{1}{4\sqrt{3}} \tan^{-1}\left(\frac{\cos x}{\sqrt{3}}\right) - \frac{1}{8} \ln\left|\frac{1+\cos x}{1-\cos x}\right| + C \]