STD 12 - 7.2 definite integral — JEE STD 11 Science — Question

$=-12\left[\left(x^2 \cdot\left(1-x^2\right)^5\right)_0^1-2 \int_0^1 x \cdot\left(1-x^2\right)^5 d x\right]=12 \int_0^1 2 x\left(1-x^2\right)^5 d x=-12 \int_1^0 t^5 d t=\frac{12}{6}\left(t^8\right)_0^1=2 .$

 Alternative : 

$\int_0^1 4 x^3\left\{\frac{d^2}{d x^2}\left(1-x^2\right)^5\right\} d x $

$\frac{d}{d x}\left(\frac{d\left(1^{\star}-x^2\right)^5}{d x}\right)=\frac{d}{d x}\left(5\left(1-x^2\right)^4(-3 x)\right)$

$=-10 \frac{d}{d x}\left(x\left(1-x^2\right)^4\right) $

$=-10\left[\left(1-x^2\right)^4+x^4\left(1-x^2\right)^3(-2 x)\right] $

$=\left[-10\left(1-x^2\right)^3\left[1-x^2-8 x^2\right]\right.$

Hence Integral

$=-40 \int_0^1 x^3\left(1-x^2\right)^3\left(1-9 x^2\right) d x \quad \quad \text { Put } x=\sin \theta$

$=-40 \int_0^{\pi / 2} \sin ^3 \theta \cos ^7 \theta d \theta+360 \int_0^1 \sin ^5 \theta \cos ^7 \theta d \theta $

$=-40.1 \cdot \frac{2.6 .4 .2}{10.8 .6 .4 .2}+360 \cdot \frac{4.2 .6 \cdot 4.2}{12.10 .8 .6 .4 .2}=-1+3=2$