If `f(x)=(sin^(-1)x)/(sqrt(1-x^(2))) and g(x)=e^(sin^(-1)x),` then `int f(x)g(x)dx` is equal to

Options

A. `e^(sin^(-1)x)(sin^(-1)x-1)+C`
B. `e^(sin^(-1)x)+C`
C. `e^((sin^(-1)x)^(2))+C`
D. `e^(2sin^(-1)x)+C`

Explanation

To solve the integral \( \int f(x) g(x) \, dx \), where \( f(x) = \frac{\sin^{-1} x}{\sqrt{1 - x^2}} \) and \( g(x) = e^{\sin^{-1} x} \), we can follow these steps:

### Step 1: Substitute \( t = \sin^{-1} x \)
Let \( t = \sin^{-1} x \). Then, we differentiate both sides:
\[ \frac{dt}{dx} = \frac{1}{\sqrt{1 - x^2}} \implies dx = \sqrt{1 - x^2} \, dt \]
This means \( \sqrt{1 - x^2} = \cos(t) \) since \( \sin^2(t) + \cos^2(t) = 1 \).

### Step 2: Rewrite the integral
Substituting \( t \) into the integral, we have:
\[ \int f(x) g(x) \, dx = \int \frac{t}{\cos(t)} e^t \cos(t) \, dt \]
The \( \cos(t) \) terms cancel out:
\[ = \int t e^t \, dt \]

### Step 3: Use integration by parts
To solve \( \int t e^t \, dt \), we apply integration by parts. Let:
- \( u = t \) and \( dv = e^t \, dt \)
Then, we find \( du = dt \) and \( v = e^t \).

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ \int t e^t \, dt = t e^t - \int e^t \, dt \]
Calculating the remaining integral:
\[ = t e^t - e^t + C \]

### Step 4: Substitute back for \( t \)
Now, we substitute back \( t = \sin^{-1} x \):
\[ = \sin^{-1} x \cdot e^{\sin^{-1} x} - e^{\sin^{-1} x} + C \]
Factoring out \( e^{\sin^{-1} x} \):
\[ = e^{\sin^{-1} x} (\sin^{-1} x - 1) + C \]

### Final Answer
Thus, the integral \( \int f(x) g(x) \, dx \) is:
\[ \int f(x) g(x) \, dx = e^{\sin^{-1} x} (\sin^{-1} x - 1) + C \]

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