Evaluation of $\int_{0}^{1}4x^3\cdot \left\{\frac{d^2}{dx^2}\left(1-x^2\right)^5\right\}dx$ The value of $\displaystyle \int_{0}^{1}4x^3\cdot \left\{\frac{d^2}{dx^2}\left(1-x^2\right)^5\right\}dx = $
$\bf{My\; Try::}$ Let $\displaystyle I = \int 4x^3\cdot \left\{\frac{d^2}{dx^2}\left(1-x^2\right)^5\right\}dx$
Using Integration by parts, we get
$\displaystyle I = 4x^3\cdot \int\left\{\frac{d^2}{dx^2}\left(1-x^2\right)^5\right\}dx - 12\int\left\{x^2\cdot \int\left\{\frac{d^2}{dx^2}\left(1-x^2\right)^5\right\}dx\right\}dx$
$\displaystyle I = 4x^3\cdot \frac{d}{dx}(1-x^2)^5-12\int \left\{x^2\cdot \frac{d}{dx}(1-x^2)^5\right\}dx$
$\displaystyle I = 4x^3\cdot 5\cdot (1-x^2)^4\cdot -2x+120\int x^2\cdot (1-x^2)^4\cdot xdx$
Now Let $x^2=t\;,$ Then $2xdx = dt$
$\displaystyle I = -10x^4\cdot (1-x^2)^4+60\int t\cdot (1-t)^4dt=-10x^4\cdot (1-x^2)^4+60\int (1-t)t^4dt$
So $\displaystyle I = -10x^4\cdot (1-x^2)^4+60\left\{\frac{t^5}{5}-\frac{t^6}{6}\right\}+\mathbb{C}$
Is my process is right?,If not then how can we solve it.
Help Required.
Thanks
 A: I wonder if it would not be simpler to establish that $$\frac{d^2}{dx^2}\left(1-x^2\right)^5=80 x^2 \left(1-x^2\right)^3-10 \left(1-x^2\right)^4=-10 \left(x^2-1\right)^3 \left(9 x^2-1\right)$$ and to use the fact that $4x^3=2x^2\frac{d(x^2)}{dx}$ which makes the change of variable $x^2=t$ quite clear. Then expansion of the integrand is quite simple in terms of $t$ and integration quite easy.
A: This appeared in my test paper today i.e JEE-Advanced 2014, I am curious as to how you got your hands on this one. :P
As for the solution, apply IBP once to get (I am dropping the factor of 4 for the moment):
$$I=\left(x^3 \frac{d}{dx}(1-x^2)^5 \right|_0^1-\int_0^1 3x^2\frac{d}{dx}(1-x^2)^5\,dx$$
Notice that the first term is zero, hence,
$$I= 15\int_0^1 x^2(1-x^2)^4 (2x)\,dx$$
With the substitution $x^2=t$, you get:
$$I=15\int_0^1 t(1-t)^4\,dt=15 \int_0^1 (1-t)t^4\,dt=15\left(\frac{t^5}{5}-\frac{t^6}{6}\right|_0^1$$
$$\Rightarrow I=\frac{1}{2}$$
Multiplying the result by 4, our final answer is: $\boxed{2}$
