`int tan(sin^(-1)x)dx` is equal to

Options

A. `(1)/(sqrt(1-x^(2)))+C`
B. `sqrt(1-x^(2))+C`
C. `(-x)/(sqrt(1-x^(2)))+C`
D. `(x)/(sqrt(1-x^(2)))+C`

Explanation

To solve the integral \( \int \tan(\sin^{-1} x) \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
Let \( I = \int \tan(\sin^{-1} x) \, dx \).

### Step 2: Use the Identity for Tangent
Using the identity \( \tan(\sin^{-1} x) = \frac{x}{\sqrt{1 - x^2}} \), we can rewrite the integral:
\[ I = \int \frac{x}{\sqrt{1 - x^2}} \, dx \]

### Step 3: Substitution
Let \( t = 1 - x^2 \). Then, differentiating gives:
\[ dt = -2x \, dx \quad \Rightarrow \quad x \, dx = -\frac{1}{2} dt \]
Now, we need to express \( \sqrt{1 - x^2} \) in terms of \( t \):
\[ \sqrt{1 - x^2} = \sqrt{t} \]

### Step 4: Substitute in the Integral
Substituting these into the integral, we have:
\[ I = \int \frac{x}{\sqrt{1 - x^2}} \, dx = \int \frac{-\frac{1}{2} dt}{\sqrt{t}} = -\frac{1}{2} \int t^{-1/2} \, dt \]

### Step 5: Integrate
Now we can integrate:
\[ -\frac{1}{2} \int t^{-1/2} \, dt = -\frac{1}{2} \cdot 2 t^{1/2} + C = -\sqrt{t} + C \]

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

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

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