If a1, a2 and a3 are the three values of a which satisfy the equation ∫0π/2(sin x + a cos x)3 dx − 4aπ − 2 ∫0π/2x cos x dx = 2 then find the value of (a12 + a22 + a32).

∫0π/2(sin x + α cos x)3 dx − 4απ − 2 ∫0π/2x cos x dx = 2Let I1 = ∫0π/2(sin x + α cos x)3 dx = ∫0π/2(sin3 x + α3 cos3 x + 3α sin2 x cos x + 3α2 sin x cos2 x)3 dx = ∫0π/2sin3 x dx + α3 ∫0π/2cos3 x dx + 3α ∫0π/2sin2 x cos x dx + 3α2 ∫0π/2sin x cos2 x dx= 23 + α3 (23) + 3α ∫01t2 dt + 3α2 ∫01t2 dt (sin x = t ⇒ cos x dx = dt; ∫01t2 dt) = 23(1 + α3) + α + α2 = 23 + 2α33 + α + α2 I1 = 2α33 + α2 + α + 23 I2 = ∫0π/2x⏟I ⋅ cos x⏟II dx = [x sin x]0π/2 − ∫0π/2sin x dx I2 = π2 − 1 = π − 22 I = 2α33 + α2 + α + 23 − 4απ − 2 − π− 22 ⇒ 2α33 + α2 − α + 23 = 2 ⇒ 2α3 + 3α3 − 3α + 2 = 6⇒ 2α3 + 3α3 − 3α − 4 = 0 α1 + α2 + α3 = 32⇒ ∑α1 α2 = − 32∴∑a12=94+62=214∴1000∑a12=1000×214=250×21=5250

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

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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 $I_{1} = \int_{0}^{π / 2} {(\sin x + α \cos x)}^{3} d x$
$= \int_{0}^{π / 2} {(\sin^{3} x + α^{3} \cos^{3} x + 3 α \sin^{2} x \cos x + 3 α^{2} \sin x \cos^{2} x)}^{3} d x$

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

$= \frac{2}{3} (1 + α^{3}) + α + α^{2} = \frac{2}{3} + \frac{2 α^{3}}{3} + α + α^{2}$

$I_{1} = \frac{2 α^{3}}{3} + α^{2} + α + \frac{2}{3}$

$I_{2} = \int_{0}^{π / 2} \underset{I}{\underset{⏟}{x}} \cdot \underset{I I}{\underset{⏟}{\cos x}} d x = {[x \sin x]}_{0}^{π / 2} - \int_{0}^{π / 2} \sin x d x$
$I_{2} = \frac{π}{2} - 1 = \frac{π - 2}{2}$

$I = \frac{2 α^{3}}{3} + α^{2} + α + \frac{2}{3} - \frac{4 α}{π - 2} - \frac{π - 2}{2}$

$\Rightarrow \frac{2 α^{3}}{3} + α^{2} - α + \frac{2}{3} = 2$
$\Rightarrow 2 α^{3} + 3 α^{3} - 3 α + 2 = 6$
$\Rightarrow 2 α^{3} + 3 α^{3} - 3 α - 4 = 0$