The equation of a curve passing through origin is given by `y = int x^(3) cos x^(4) dx ` . If the equation of the curve is wirtten in the form x = g(y) , then

Options

A. `g(y) root(3)(sin^(-1) (4y))`
B. `g(y) = sqrt(sin^(-1)(4y))`
C. `g(y) = root(4)(sin^(-1)(4y))`
D. None of these

Answer

To solve the problem step by step, we will start from the given equation of the curve and perform the necessary integrations and substitutions.

### Step 1: Set up the integral
We are given the equation of the curve:
\[ y = \int x^3 \cos(x^4) \, dx \]

### Step 2: Use substitution for integration
Let’s use the substitution method. We will let:
\[ t = x^4 \]
Then, differentiating both sides gives:
\[ dt = 4x^3 \, dx \]
This implies:
\[ dx = \frac{dt}{4x^3} \]

### Step 3: Substitute in the integral
Now, we can express \( x^3 \, dx \) in terms of \( dt \):
\[ x^3 \, dx = \frac{1}{4} dt \]
Substituting this into the integral, we get:
\[ y = \int x^3 \cos(x^4) \, dx = \int \cos(t) \cdot \frac{1}{4} dt \]
This simplifies to:
\[ y = \frac{1}{4} \int \cos(t) \, dt \]

### Step 4: Integrate
The integral of \( \cos(t) \) is \( \sin(t) \), so we have:
\[ y = \frac{1}{4} \sin(t) + C \]
Substituting back \( t = x^4 \):
\[ y = \frac{1}{4} \sin(x^4) + C \]

### Step 5: Determine the constant of integration
Since the curve passes through the origin (0, 0), we can find \( C \):
When \( x = 0 \), \( y = 0 \):
\[ 0 = \frac{1}{4} \sin(0) + C \]
This implies:
\[ C = 0 \]
Thus, the equation simplifies to:
\[ y = \frac{1}{4} \sin(x^4) \]

### Step 6: Rearranging to find \( x \) in terms of \( y \)
Now we want to express \( x \) in terms of \( y \). Rearranging the equation:
\[ 4y = \sin(x^4) \]
Taking the inverse sine:
\[ x^4 = \sin^{-1}(4y) \]
Now, taking the fourth root gives:
\[ x = ( \sin^{-1}(4y) )^{1/4} \]

### Final Result
Thus, we can express the equation in the form \( x = g(y) \):
\[ g(y) = ( \sin^{-1}(4y) )^{1/4} \]