If α1,α2 and α3are the three values of a which satisfy the equation ∫0π/2(sinx+αcosx)3dx−4απ−2∫0π/2xcosxdx=2 then the value of 1000(α12+α22+α32) is__________

∫0π/2(sinx+αcosx)3dx−4απ−2∫0π/2xcosxdx=2Let VI1=∫0π/2(sinx+αcosx)3dx =∫0π/2(sin3x+α3cos3x+3αsin2xcosx+3α2sinxcos2x)dx+3α2∫0π/2sinxcos2xdx(sinx=t⇒cosxdx=dt;∫01t2dt) =23+α3(23)+3α∫01t2dt+3α2∫01t2dt =23(1+α3)+α+α2 =23+2α33+α+α2 I1=2α33+α2+α+23 I2=∫0π/2x⏟(I).cosx⏟(II)dx=xsinx|0π/2−∫0π/2sinxdxI2=xsinx+cosx|0π/2 I2=π2−1=π−22 I=2α33+α2+α+23−4απ−2.π−22 ⇒2α33+α2−α+23=2 ⇒2α3+3α2−3α+2=6⇒2α3+3α2−3α−4=0 α1+α2+α3=−32 ⇒∑α1α2=−32 ∴∑α12=94+62=214∴1000∑α12=1000×214=250×21=5250

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If $α_{1}, α_{2}$ and $α_{3}$ are the three values of a which satisfy the equation $\int_{0}^{π / 2} {(\sin x + α \cos x)}^{3} d x - \frac{4 α}{π - 2} \int_{0}^{π / 2} x \cos x d x = 2$ then the value of $1000 (α_{1}^{2} + α_{2}^{2} + α_{3}^{2})$ is__________

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Detailed Solution

$\int_{0}^{π / 2} {(\sin x + α \cos x)}^{3} d x - \frac{4 α}{π - 2} \int_{0}^{π / 2} x \cos x d x = 2$
Let $V I_{1} = \int_{0}^{π / 2} {(\sin x + α \cos x)}^{3} d x$
$= \int_{0}^{π / 2} (\sin^{3} x + α^{3} \cos^{3} x + 3 α \sin^{2} x \cos x + 3 α^{2} \sin x \cos^{2} x) d x + 3 α^{2} \int_{0}^{π / 2} \sin x \cos^{2} x d x$ $(\sin x = t \Rightarrow \cos x d x = d t; \int_{0}^{1} t^{2} d t)$
$= \frac{2}{3} + α^{3} (\frac{2}{3}) + 3 α \int_{0}^{1} t^{2} d t + 3 α^{2} \int_{0}^{1} t^{2} d t$
$= \frac{2}{3} (1 + α^{3}) + α + α^{2}$
$= \frac{2}{3} + \frac{2 α^{3}}{3} + α + α^{2}$

$I_{1} = \frac{2 α^{3}}{3} + α^{2} + α + \frac{2}{3}$
$I_{2} = \int_{0}^{π / 2} \underset{(I)}{\underset{⏟}{x}} . \underset{(I I)}{\underset{⏟}{\cos x}} d x = x \sin x |_{0}^{π / 2} - \int_{0}^{π / 2} \sin x d x$ $I_{2} = x \sin x + \cos {x |}_{0}^{π / 2}$

$I_{2} = \frac{π}{2} - 1 = \frac{π - 2}{2}$

$I = \frac{2 α^{3}}{3} + α^{2} + α + \frac{2}{3} - \frac{4 α}{π - 2} . \frac{π - 2}{2}$
$\Rightarrow \frac{2 α^{3}}{3} + α^{2} - α + \frac{2}{3} = 2$
$\Rightarrow 2 α^{3} + 3 α^{2} - 3 α + 2 = 6$
$\Rightarrow 2 α^{3} + 3 α^{2} - 3 α - 4 = 0$
$α_{1} + α_{2} + α_{3} = - \frac{3}{2}$
$\Rightarrow \sum α_{1} α_{2} = - \frac{3}{2}$
$∴ \sum α_{1}^{2} = \frac{9}{4} + \frac{6}{2} = \frac{21}{4}$
$∴ 1000 \sum α_{1}^{2} = 1000 \times \frac{21}{4} = 250 \times 21 = 5250$