If `int f(x)dx=F(x),` then `intx^3f(x^2)dx` is equal to :

Options

A. `1/2[x^(2){F(x)}^(2)dx]`
B. `1/2[x^(2)F(x^(2))-intF(x^(2))d(x^(2))]`
C. `1/2[x^(2)F(x)-1/2 int {F(x)}^(2)dx]`
D. None of the above

Answer

To solve the integral \( \int x^3 f(x^2) \, dx \), we can use a substitution method. Here’s a step-by-step solution:

### Step 1: Substitution
Let \( t = x^2 \). Then, we differentiate to find \( dx \):
\[ dt = 2x \, dx \implies dx = \frac{dt}{2x} \]
Since \( x = \sqrt{t} \), we can substitute \( dx \) in terms of \( t \):
\[ dx = \frac{dt}{2\sqrt{t}} \]

### Step 2: Change the Integral
Now we substitute \( x^3 \) and \( f(x^2) \) in the integral:
\[ x^3 = (x^2)^{3/2} = t^{3/2} \quad \text{and} \quad f(x^2) = f(t) \]
Thus, the integral becomes:
\[ \int x^3 f(x^2) \, dx = \int t^{3/2} f(t) \cdot \frac{dt}{2\sqrt{t}} = \int \frac{t^{3/2}}{2\sqrt{t}} f(t) \, dt \]
Simplifying \( \frac{t^{3/2}}{2\sqrt{t}} \) gives:
\[ \int \frac{t^{3/2}}{2\sqrt{t}} f(t) \, dt = \int \frac{t^{3/2}}{2t^{1/2}} f(t) \, dt = \int \frac{t}{2} f(t) \, dt \]

### Step 3: Factor out the constant
Now we can factor out the constant \( \frac{1}{2} \):
\[ = \frac{1}{2} \int t f(t) \, dt \]

### Step 4: Integration by Parts
To solve \( \int t f(t) \, dt \), we can use integration by parts:
Let \( u = t \) and \( dv = f(t) \, dt \). Then, we have:
\[ du = dt \quad \text{and} \quad v = F(t) \quad \text{(since \( \int f(t) \, dt = F(t) \))} \]
Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
We get:
\[ \int t f(t) \, dt = t F(t) - \int F(t) \, dt \]

### Step 5: Substitute back
Now substituting back into our integral:
\[ \frac{1}{2} \left( t F(t) - \int F(t) \, dt \right) \]
Substituting \( t = x^2 \):
\[ = \frac{1}{2} \left( x^2 F(x^2) - \int F(x^2) \, dt \right) \]

### Final Result
Thus, the final result is:
\[ \int x^3 f(x^2) \, dx = \frac{1}{2} \left( x^2 F(x^2) - \int F(x^2) \, dt \right) \]