`int (dx)/((4sin^(2)x+5cos^(2)x))=?`

Options

A. `(1)/(2)tan^(-1)((tanx)/(sqrt(5)))+C`
B. `(1)/(sqrt(5))tan^(-1)((tanx)/(sqrt(5)))+C`
C. `(1)/(2sqrt(5))tan^(-1)((2tanx)/(sqrt(5)))+C`
D. None of these

Explanation

To solve the integral \( \int \frac{dx}{4 \sin^2 x + 5 \cos^2 x} \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{dx}{4 \sin^2 x + 5 \cos^2 x} \]

### Step 2: Divide by \( \cos^2 x \)
To simplify the expression, we divide the numerator and denominator by \( \cos^2 x \):
\[ I = \int \frac{\frac{1}{\cos^2 x}}{\frac{4 \sin^2 x}{\cos^2 x} + 5} \, dx \]
This simplifies to:
\[ I = \int \frac{\sec^2 x}{4 \tan^2 x + 5} \, dx \]

### Step 3: Substitute \( t = \tan x \)
Let \( t = \tan x \). Then, the derivative \( dt = \sec^2 x \, dx \) or \( dx = \frac{dt}{\sec^2 x} \). Thus, we can rewrite the integral as:
\[ I = \int \frac{dt}{4t^2 + 5} \]

### Step 4: Factor out the Constant
We can factor out the constant from the denominator:
\[ I = \int \frac{dt}{4(t^2 + \frac{5}{4})} \]
This can be simplified to:
\[ I = \frac{1}{4} \int \frac{dt}{t^2 + \frac{5}{4}} \]

### Step 5: Use the Standard Integral Formula
The integral \( \int \frac{dt}{t^2 + a^2} = \frac{1}{a} \tan^{-1} \left( \frac{t}{a} \right) + C \) can be applied here, where \( a = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2} \):
\[ I = \frac{1}{4} \cdot \frac{2}{\sqrt{5}} \tan^{-1} \left( \frac{t}{\frac{\sqrt{5}}{2}} \right) + C \]

### Step 6: Substitute Back for \( t \)
Now we substitute back \( t = \tan x \):
\[ I = \frac{1}{2\sqrt{5}} \tan^{-1} \left( \frac{2 \tan x}{\sqrt{5}} \right) + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{dx}{4 \sin^2 x + 5 \cos^2 x} = \frac{1}{2\sqrt{5}} \tan^{-1} \left( \frac{2 \tan x}{\sqrt{5}} \right) + C \]