`int(sinxcosx)/(sqrt(1-sin^(4)x)dx` is equal to

Options

A. `(1)/(2) sin^(-1) (sin^2 x)+C`
B. `-(1)/(2)cos^(-1)(sin^(2)x)+C`
C. `tan^(-1)(sin^(2)x) + C`
D. `cot^(-1) (sin x)+C`

Answer

To solve the integral \( \int \frac{\sin x \cos x}{\sqrt{1 - \sin^4 x}} \, dx \), we can follow these steps:

### Step 1: Substitution
Let \( t = \sin^2 x \). Then, we differentiate \( t \):
\[ dt = 2 \sin x \cos x \, dx \quad \Rightarrow \quad dx = \frac{dt}{2 \sin x \cos x} \]

### Step 2: Rewrite the Integral
Now we can substitute \( t \) and \( dx \) into the integral:
\[ \int \frac{\sin x \cos x}{\sqrt{1 - \sin^4 x}} \, dx = \int \frac{\sin x \cos x}{\sqrt{1 - t^2}} \cdot \frac{dt}{2 \sin x \cos x} \]
The \( \sin x \cos x \) terms cancel out:
\[ = \frac{1}{2} \int \frac{dt}{\sqrt{1 - t^2}} \]

### Step 3: Integrate
The integral \( \int \frac{dt}{\sqrt{1 - t^2}} \) is a standard integral that equals \( \sin^{-1}(t) \):
\[ = \frac{1}{2} \sin^{-1}(t) + C \]

### Step 4: Substitute Back
Now substitute back \( t = \sin^2 x \):
\[ = \frac{1}{2} \sin^{-1}(\sin^2 x) + C \]

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