Method: Integration by Parts

The given integral is: $I = \int\limits_0^1 {4{x^3}\left\{ {{{{d^2}} \over {d{x^2}}}{{\left( {1 - {x^2}} \right)}^5}} \right\}dx}$ This integral is in the form $\int u \, dv$, which is suitable for integration by parts. The formula for definite integration by parts is $\int_a^b u \, dv = [uv]_a^b - \int_a^b v \, du$.

We will apply integration by parts twice.

Step 1: First Integration by Parts

Let's choose the parts for the integral:

• Let $u = 4x^3$.
• Let $dv = \frac{d^2}{dx^2}(1-x^2)^5 dx$.

Now, we find $du$ and $v$:

• $du = \frac{d}{dx}(4x^3) dx = 12x^2 dx$.
• $v = \int \frac{d^2}{dx^2}(1-x^2)^5 dx = \frac{d}{dx}(1-x^2)^5$.

Applying the integration by parts formula: $I = \left[ 4x^3 \cdot \frac{d}{dx}(1-x^2)^5 \right]_0^1 - \int_0^1 \left( \frac{d}{dx}(1-x^2)^5 \right) (12x^2) dx$ First, let's evaluate the boundary term. We need the first derivative of $(1-x^2)^5$: $\frac{d}{dx}(1-x^2)^5 = 5(1-x^2)^4(-2x) = -10x(1-x^2)^4$ So the boundary term is: $\left[ 4x^3 (-10x(1-x^2)^4) \right]_0^1 = \left[ -40x^4(1-x^2)^4 \right]_0^1$

• At $x=1$: $-40(1)^4(1-1^2)^4 = -40(1)(0) = 0$.
• At $x=0$: $-40(0)^4(1-0^2)^4 = 0$. Since the boundary term evaluates to $0 - 0 = 0$, the integral simplifies to: $I = - \int_0^1 \left( \frac{d}{dx}(1-x^2)^5 \right) (12x^2) dx$

Step 2: Second Integration by Parts

Now we need to evaluate the new integral: $I = -12 \int_0^1 x^2 \frac{d}{dx}(1-x^2)^5 dx$ Let's apply integration by parts again to the integral $\int_0^1 x^2 \frac{d}{dx}(1-x^2)^5 dx$.

• Let $u_1 = x^2$.
• Let $dv_1 = \frac{d}{dx}(1-x^2)^5 dx$.

Then:

• $du_1 = 2x dx$.
• $v_1 = (1-x^2)^5$.

Applying the formula again: $\int_0^1 x^2 \frac{d}{dx}(1-x^2)^5 dx = \left[ x^2(1-x^2)^5 \right]_0^1 - \int_0^1 (1-x^2)^5 (2x) dx$ Evaluate the boundary term:

• At $x=1$: $1^2(1-1^2)^5 = 1(0) = 0$.
• At $x=0$: $0^2(1-0^2)^5 = 0(1) = 0$. The boundary term is again $0 - 0 = 0$. So, $\int_0^1 x^2 \frac{d}{dx}(1-x^2)^5 dx = - \int_0^1 2x(1-x^2)^5 dx$ Substituting this back into our expression for $I$: $I = -12 \left( - \int_0^1 2x(1-x^2)^5 dx \right) = 24 \int_0^1 x(1-x^2)^5 dx$

Step 3: Solving the Final Integral

We now have a much simpler integral to solve: $I = 24 \int_0^1 x(1-x^2)^5 dx$ Let's use the substitution method.

• Let $t = 1-x^2$.
• Then $dt = -2x dx$, which implies $x dx = -\frac{dt}{2}$.

We also need to change the limits of integration:

• When $x=0$, $t = 1 - 0^2 = 1$.
• When $x=1$, $t = 1 - 1^2 = 0$.

Substituting these into the integral: $I = 24 \int_1^0 t^5 \left(-\frac{dt}{2}\right)$ $I = -12 \int_1^0 t^5 dt$ We can reverse the limits of integration, which changes the sign of the integral: $I = 12 \int_0^1 t^5 dt$ Now, we can perform the integration: $I = 12 \left[ \frac{t^6}{6} \right]_0^1$ $I = 12 \left( \frac{1^6}{6} - \frac{0^6}{6} \right)$ $I = 12 \left( \frac{1}{6} \right)$ $I = 2$

Thus, the value of the integral is 2.