If [tex]$x$[/tex] satisfies the equation [tex]$\left(\int_0^1 \frac{d t}{t^2+2 t \cos \alpha+1}\right) x^2$ $-\left(\int_{-3}^3 \frac{t^2 \sin 2 t}{t^2+1} d t\right) x-2=0(0\ \textless \ \alpha\ \textless \ \pi)$[/tex], then the value of [tex]$x$[/tex] is
(a) [tex]$\pm \sqrt{\frac{\alpha}{2 \sin \alpha}}$[/tex]
(b) [tex]$\pm \sqrt{\frac{2 \sin \alpha}{\alpha}}$[/tex]
(c) [tex]$\pm \sqrt{\frac{\alpha}{\sin \alpha}}$[/tex]
(d) [tex]$\pm 2 \sqrt{\frac{\sin \alpha}{\alpha}}$[/tex]
Answer (1)
Evaluate the integral ∫ − 3 3 t 2 + 1 t 2 s i n 2 t d t , recognizing it as the integral of an odd function over a symmetric interval, which equals 0.
Evaluate the integral ∫ 0 1 t 2 + 2 t c o s α + 1 d t by completing the square and using trigonometric substitution, resulting in 2 s i n α α .
Substitute the results into the given equation, simplifying it to ( 2 s i n α α ) x 2 − 2 = 0 .
Solve for x , obtaining x = ± 2 α s i n α .
± 2 α sin α
Explanation
Problem Setup We are given the equation
( ∫ 0 1 t 2 + 2 t c o s α + 1 d t ) x 2 − ( ∫ − 3 3 t 2 + 1 t 2 s i n 2 t d t ) x − 2 = 0
where 0 < α < π , and we want to find the value of x .
Evaluating the Second Integral First, let's evaluate the integral ∫ − 3 3 t 2 + 1 t 2 s i n 2 t d t . Notice that the integrand is an odd function, since
f ( − t ) = ( − t ) 2 + 1 ( − t ) 2 s i n ( 2 ( − t )) = t 2 + 1 t 2 ( − s i n ( 2 t )) = − t 2 + 1 t 2 s i n ( 2 t ) = − f ( t ) .
Therefore, the integral of an odd function over a symmetric interval is zero:
∫ − 3 3 t 2 + 1 t 2 s i n 2 t d t = 0 .
Evaluating the First Integral Now, let's evaluate the integral ∫ 0 1 t 2 + 2 t c o s α + 1 d t . We can complete the square in the denominator:
t 2 + 2 t cos α + 1 = ( t + cos α ) 2 + 1 − cos 2 α = ( t + cos α ) 2 + sin 2 α .
So, the integral becomes
∫ 0 1 ( t + c o s α ) 2 + s i n 2 α d t = s i n α 1 arctan ( s i n α t + c o s α ) 0 1 = s i n α 1 [ arctan ( s i n α 1 + c o s α ) − arctan ( s i n α c o s α ) ] .
Simplifying the Integral Using the trigonometric identities s i n α 1 + c o s α = cot ( 2 α ) and s i n α c o s α = cot ( α ) , we have
s i n α 1 [ arctan ( cot ( 2 α ) ) − arctan ( cot ( α ) ) ] = s i n α 1 [ 2 π − 2 α − ( 2 π − α ) ] = s i n α 1 [ 2 α ] = 2 s i n α α .
Solving for x Substituting these results back into the original equation, we get
( 2 s i n α α ) x 2 − ( 0 ) x − 2 = 0 .
So, 2 s i n α α x 2 = 2 , which means x 2 = α 4 s i n α .
Therefore, x = ± α 4 s i n α = ± 2 α s i n α .
Final Answer Thus, the value of x is ± 2 α s i n α .
Examples
Imagine you are designing a speaker system where the acoustic response depends on an integral similar to the one in this problem. The variable x could represent the amplitude of the sound wave, and α could be related to the frequency. Solving the equation for x would allow you to determine the range of amplitudes for a given frequency to achieve optimal sound quality. This type of calculation is crucial in audio engineering to ensure the speaker system performs as desired.
