Question

If $a_1,a_2$ and $a_3$ are the three values of $a$ which satisfy the equation $\underset{0}{\overset{\pi /2}{\int }}(\sin x+a\cos x{)}^{3}dx-\frac{4a}{\pi -2}\underset{0}{\overset{\pi /2}{\int }}x\cos xdx=2,$ then the value of $1000(a_1^2+a_2^2+a_3^2)$ is

Answer 1

$I=\underset{0}{\overset{\pi /2}{\int }}(\sin x+a\cos x{)}^{3}dx-\frac{4a}{\pi -2}\underset{0}{\overset{\pi /2}{\int }}x\cos xdx=2$

Let $I_1=\underset{0}{\overset{\pi /2}{\int }}(\sin x+a\cos x{)}^{3}dx$
$=\underset{0}{\overset{\pi /2}{\int }}{\left(\frac{(\sin x+a\cos x)\sqrt{1+a^2}}{\sqrt{1+a^2}}\right)}^{3}dx$
$=(1+a^2{)}^{3/2}\underset{0}{\overset{\pi /2}{\int }}(\sin (x+\alpha ){)}^{3}dx$
where $\cos \alpha =\frac{1}{\sqrt{1+a^2}}$ and $\sin \alpha =\frac{a}{\sqrt{1+a^2}}$
$I_1=(1+a^2{)}^{3/2}\underset{0}{\overset{\pi /2}{\int }}[1-{\cos }^2(x+\alpha \left)\right]\sin (x+\alpha )dx$
Put $\cos (x+\alpha )=t$
$\Rightarrow -\sin (x+\alpha )dx=dt$
$I_1=-(1+a^2{)}^{3/2}\underset{\cos \alpha }{\overset{-\sin \alpha }{\int }}(1-t^2)dt$
$=(1+a^2{)}^{3/2}{[\frac{t^3}{3}-t]}_{\cos \alpha }^{-\sin \alpha }$
$=(1+a^2{)}^{3/2}\left[\frac{1}{3}\right(-{\sin }^3\alpha -{\cos }^3\alpha )-(-\sin \alpha -\cos \alpha \left)\right]$
$=(1+a^2{)}^{3/2}\left[\right(\sin \alpha +\cos \alpha )-\frac{1}{3}(\sin \alpha +\cos \alpha \left)\right(1-\sin \alpha \cos \alpha \left)\right]$
$=(1+a^2{)}^{3/2}\frac{(\sin \alpha +\cos \alpha )}{3}[3-1+\sin \alpha \cos \alpha ]$
$=\frac{(1+a^2{)}^{3/2}}{3}\left(\frac{1+a}{\sqrt{1+a^2}}\right)(2+\frac{a}{1+a^2})$
$=\frac{1+a}{3}(2+2a^2+a)$


$I_2=\underset{0}{\overset{\pi /2}{\int }}\underset{I}{\underset{}{x}}\cdot \underset{II}{\underset{}{\cos x}}dx$
$=x\sin x{|}_0^{\pi /2}-\underset{0}{\overset{\pi /2}{\int }}\sin xdx$
$=\frac{\pi }{2}-1=\frac{\pi -2}{2}$

$\therefore I=\frac{1+a}{3}(2+2a^2+a)-\frac{4a}{\pi -2}\cdot \frac{\pi -2}{2}=2$
$\Rightarrow 2a^3+3a^2-3a-4=0$

$a_1+a_2+a_3=-\frac{3}{2}$
$\sum a_1{a}_2=-\frac{3}{2}$
$\therefore \sum a_1^2={\left(\frac{-3}{2}\right)}^2-2\left(\frac{-3}{2}\right)$
$=\frac{9}{4}+3=\frac{21}{4}$
$\therefore 1000\sum a_1^2=1000\times \frac{21}{4}$
$=250\times 21=5250$