Let `I=int(cos^(3)x)/(1+sin^(2)x)dx`, then I is equal to (where c is the constant of integration )

Options

A. `2tan^(-1)(x)+sinx+c`
B. `2tan^(-1)(sinx)-sinx+c`
C. `2tan^(-1)(x)-x+c`
D. `2tan^(-1)(sinx)+sinx+c`

Answer

To solve the integral \( I = \int \frac{\cos^3 x}{1 + \sin^2 x} \, dx \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{\cos^3 x}{1 + \sin^2 x} \, dx \]
We can express \( \cos^3 x \) as \( \cos^2 x \cdot \cos x \):
\[ I = \int \frac{\cos^2 x \cdot \cos x}{1 + \sin^2 x} \, dx \]

### Step 2: Use a Substitution
Let \( \sin x = t \). Then, the derivative \( \cos x \, dx = dt \) or \( dx = \frac{dt}{\cos x} \). We also know that \( \cos^2 x = 1 - \sin^2 x = 1 - t^2 \). Therefore, we can rewrite the integral:
\[ I = \int \frac{(1 - t^2) \cos x}{1 + t^2} \cdot \frac{dt}{\cos x} \]
This simplifies to:
\[ I = \int \frac{1 - t^2}{1 + t^2} \, dt \]

### Step 3: Split the Integral
We can split the integral into two parts:
\[ I = \int \frac{1}{1 + t^2} \, dt - \int \frac{t^2}{1 + t^2} \, dt \]

### Step 4: Integrate Each Part
1. The first integral:
\[ \int \frac{1}{1 + t^2} \, dt = \tan^{-1}(t) \]
2. The second integral can be simplified:
\[ \int \frac{t^2}{1 + t^2} \, dt = \int \left(1 - \frac{1}{1 + t^2}\right) dt = \int dt - \int \frac{1}{1 + t^2} \, dt = t - \tan^{-1}(t) \]

Putting it all together, we have:
\[ I = \tan^{-1}(t) - \left(t - \tan^{-1}(t)\right) = 2\tan^{-1}(t) - t \]

### Step 5: Substitute Back
Recall that \( t = \sin x \):
\[ I = 2\tan^{-1}(\sin x) - \sin x + C \]

### Final Answer
Thus, the integral evaluates to:
\[ I = 2\tan^{-1}(\sin x) - \sin x + C \]

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