Evaluate the definite integral:
\[
\int_{\sqrt[3]{\log 3}}^{\sqrt[3]{\log 4}} \frac{x^2 \sin(x^3)}{\sin(x^3) + \sin(\log 12 - x^3)} \, dx
\]
Answer (1)
To evaluate the definite integral ∫ 3 l o g 3 3 l o g 4 s i n ( x 3 ) + s i n ( l o g 12 − x 3 ) x 2 s i n ( x 3 ) d x , we can utilize symmetry in the function involved.
The key observation here is that the function h ( x ) = s i n ( x 3 ) + s i n ( l o g 12 − x 3 ) x 2 s i n ( x 3 ) can be paired with its counterpart by using a substitution that involves symmetry about the limits of integration.
Let's consider a substitution u = ( lo g 12 ) 1/3 − x , so then d u = − d x . Changing the limits for ( u ) gives:
When x = 3 lo g 3 , u = ( lo g 12 ) 1/3 − 3 lo g 3 .
When x = 3 lo g 4 , u = ( lo g 12 ) 1/3 − 3 lo g 4 .
Reversing the limits and accounting for the negative sign, we have:
∫ 3 l o g 3 3 l o g 4 h ( x ) d x = ∫ 3 l o g 4 3 l o g 3 sin ((( lo g 12 ) 1/3 − u ) 3 ) + sin ( lo g 12 − (( lo g 12 ) 1/3 − u ) 3 ) (( lo g 12 ) 1/3 − u ) 2 sin ((( lo g 12 ) 1/3 − u ) 3 ) ( − d u )
Using the properties of trigonometric functions and considering the symmetry, it can be shown that the integral of the original function and its transformed counterpart over the defined interval cancel each other out. Each part sums to zero due to their symmetrical placement in relation to the midpoint of the interval and their trigonometric identities.
Thus, the value of the integral is:
∫ 3 l o g 3 3 l o g 4 h ( x ) d x = 0
The beautiful symmetry comes from the periodic properties of the sine function and the clever use of substitution.
