Science:Math Exam Resources/Courses/MATH101 B/April 2024/Question 04/Solution 1

We simplify the integral by the following substitution; let $u=t^{3}$, so $du=3t^{2}dt.$ Using this substitution, the integration bounds can be rewritten in terms of $u$ as follows: $$t=-x^{2}\implies u=-x^{6},\;{\rm {and}}\;t=x^{2}\implies u=x^{6}.$$ Then we have

$${\begin{aligned}\int _{-x^{2}}^{x^{2}}t^{2}\cos(t^{3})dt&=\int _{-x^{6}}^{x^{6}}{\frac {1}{3}}\cos(u)du\\&={\frac {1}{3}}\int _{-x^{6}}^{x^{6}}\cos(u)du\\&={\frac {1}{3}}\sin(u)|_{-x^{6}}^{x^{6}}\\&={\frac {1}{3}}\left(\sin(x^{6})-\sin(-x^{6})\right)\\&={\frac {2}{3}}\sin(x^{6})\\\end{aligned}}$$

Note that there is no constant of integration in a definite integral.