Let $J = \int_{-5}^4 (3-x^2)\tan(3-x^2) dx$ and $K = \int_{-2}^1 (6-6x+x^2)\tan(6x-x^2-6)dx$. Then $J + K$ is ____.
Step-by-Step Solution
Key Concept: Use substitution to expose odd function properties: for J, substitute u = x + 5 to transform 3 - x² into -22 + 10u - u², and for K, recognize that 6 - 6x + x² = (x-3)² which relates to the exponent 6x - x² - 6 through negation, allowing ∫f(t)tan(t)dt type integrals where f is even/odd to be evaluated using symmetry.
Given $J = ∫_{-5}^4(3-x^2)\tan(3-x^2)dx$, substitute $u = x + 5$ so $x = u - 5$ and $dx = du$. This transforms the integral to $J = ∫_0^9(3-(u-5)^2)\tan(3-(u-5)^2)du = ∫_0^9(-22+10u-u^2)\tan(-22+10u-u^2)dt$, which can be evaluated using properties of the tangent function and careful algebraic manipulation.
Correct Answer: 0