Question

If r1,r2 and r3 are three values of r which satisfy the equation ∫π/20(sinx+rcosx)3dx−4rπ−2∫π/20xcosxdx=2 then the value of 47(r21+r22+r23) is equal to

Answer 1

47(12π - 3/4)

Answer 2

Let the given equation be

$$\int_{0}^{\pi/2} (\sin x + r \cos x)^3 dx - 4r\pi - 2 \int_{0}^{\pi/2} x \cos x dx = 2$$

First, evaluate the integral $I_1 = \int_{0}^{\pi/2} (\sin x + r \cos x)^3 dx$.
Expand the integrand:
$(\sin x + r \cos x)^3 = \sin^3 x + 3 \sin^2 x (r \cos x) + 3 \sin x (r \cos x)^2 + (r \cos x)^3$
$= \sin^3 x + 3r \sin^2 x \cos x + 3r^2 \sin x \cos^2 x + r^3 \cos^3 x$

Integrate term by term from 0 to $\pi/2$:
$\int_{0}^{\pi/2} \sin^3 x dx = \int_{0}^{\pi/2} (1-\cos^2 x) \sin x dx$. Let $u = \cos x$, $du = -\sin x dx$.
When $x=0, u=1$. When $x=\pi/2, u=0$.
$\int_{1}^{0} (1-u^2)(-du) = \int_{0}^{1} (1-u^2)du = [u - \frac{u^3}{3}]_{0}^{1} = 1 - \frac{1}{3} = \frac{2}{3}$.

$\int_{0}^{\pi/2} 3r \sin^2 x \cos x dx$. Let $u = \sin x$, $du = \cos x dx$.
When $x=0, u=0$. When $x=\pi/2, u=1$.
$\int_{0}^{1} 3r u^2 du = [r u^3]_{0}^{1} = r(1^3 - 0^3) = r$.

$\int_{0}^{\pi/2} 3r^2 \sin x \cos^2 x dx$. Let $u = \cos x$, $du = -\sin x dx$.
When $x=0, u=1$. When $x=\pi/2, u=0$.
$\int_{1}^{0} 3r^2 u^2 (-du) = \int_{0}^{1} 3r^2 u^2 du = [r^2 u^3]_{0}^{1} = r^2(1^3 - 0^3) = r^2$.

$\int_{0}^{\pi/2} r^3 \cos^3 x dx = \int_{0}^{\pi/2} r^3 (1-\sin^2 x) \cos x dx$. Let $u = \sin x$, $du = \cos x dx$.
When $x=0, u=0$. When $x=\pi/2, u=1$.
$\int_{0}^{1} r^3 (1-u^2) du = r^3 [u - \frac{u^3}{3}]_{0}^{1} = r^3 (1 - \frac{1}{3}) = \frac{2}{3} r^3$.

So, $I_1 = \frac{2}{3} + r + r^2 + \frac{2}{3} r^3$.

Next, evaluate the integral $I_2 = \int_{0}^{\pi/2} x \cos x dx$ using integration by parts, $\int u dv = uv - \int v du$.
Let $u = x$, $dv = \cos x dx$. Then $du = dx$, $v = \sin x$.
$I_2 = [x \sin x]_{0}^{\pi/2} - \int_{0}^{\pi/2} \sin x dx$
$I_2 = (\frac{\pi}{2} \sin \frac{\pi}{2} - 0 \sin 0) - [-\cos x]_{0}^{\pi/2}$
$I_2 = (\frac{\pi}{2} \times 1 - 0) - (-\cos \frac{\pi}{2} - (-\cos 0))$
$I_2 = \frac{\pi}{2} - (0 - (-1)) = \frac{\pi}{2} - 1$.

Substitute the values of $I_1$ and $I_2$ into the given equation:
$(\frac{2}{3} + r + r^2 + \frac{2}{3} r^3) - 4r\pi - 2 (\frac{\pi}{2} - 1) = 2$
$\frac{2}{3} + r + r^2 + \frac{2}{3} r^3 - 4r\pi - (\pi - 2) = 2$
$\frac{2}{3} + r + r^2 + \frac{2}{3} r^3 - 4r\pi - \pi + 2 = 2$
$\frac{2}{3} + r + r^2 + \frac{2}{3} r^3 - 4r\pi - \pi = 0$
Multiply the entire equation by 3 to eliminate the fraction:
$2 + 3r + 3r^2 + 2r^3 - 12r\pi - 3\pi = 0$
Rearrange the terms to form a cubic equation in $r$:
$2r^3 + 3r^2 + (3 - 12\pi)r + (2 - 3\pi) = 0$.

The problem states that $r_1, r_2, r_3$ are three values of $r$ which satisfy this equation. Thus, $r_1, r_2, r_3$ are the roots of the cubic equation $2r^3 + 3r^2 + (3 - 12\pi)r + (2 - 3\pi) = 0$.

For a cubic equation $ax^3 + bx^2 + cx + d = 0$ with roots $r_1, r_2, r_3$, Vieta's formulas give the following relationships:
Sum of roots: $r_1 + r_2 + r_3 = -\frac{b}{a}$
Sum of products of roots taken two at a time: $r_1 r_2 + r_2 r_3 + r_3 r_1 = \frac{c}{a}$

In our equation, $a=2$, $b=3$, $c=3 - 12\pi$, and $d=2 - 3\pi$.
So, the sum of the roots is $r_1 + r_2 + r_3 = -\frac{3}{2}$.
The sum of the products of the roots taken two at a time is $r_1 r_2 + r_2 r_3 + r_3 r_1 = \frac{3 - 12\pi}{2}$.

We need to find the value of $r_1^2 + r_2^2 + r_3^2$. We use the identity:
$(r_1 + r_2 + r_3)^2 = r_1^2 + r_2^2 + r_3^2 + 2(r_1 r_2 + r_2 r_3 + r_3 r_1)$
Rearranging the identity to solve for the sum of squares:
$r_1^2 + r_2^2 + r_3^2 = (r_1 + r_2 + r_3)^2 - 2(r_1 r_2 + r_2 r_3 + r_3 r_1)$

Substitute the values from Vieta's formulas:
$r_1^2 + r_2^2 + r_3^2 = \left(-\frac{3}{2}\right)^2 - 2\left(\frac{3 - 12\pi}{2}\right)$
$r_1^2 + r_2^2 + r_3^2 = \frac{9}{4} - (3 - 12\pi)$
$r_1^2 + r_2^2 + r_3^2 = \frac{9}{4} - 3 + 12\pi$
$r_1^2 + r_2^2 + r_3^2 = -\frac{3}{4} + 12\pi$.

The question asks for the value of $47(r_1^2 + r_2^2 + r_3^2)$.
$47(r_1^2 + r_2^2 + r_3^2) = 47\left(-\frac{3}{4} + 12\pi\right)$
$47(r_1^2 + r_2^2 + r_3^2) = -\frac{141}{4} + 564\pi$.

Thus,

$47(r_1^2 + r_2^2 + r_3^2) = 47(12\pi - \frac{3}{4})$