Science:Math Exam Resources/Courses/MATH101 A/April 2024/Question 04

MATH101 A April 2024

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[hide] Question 04
Evaluate $\int _{-x^{2}}^{x^{2}}t^{2}\cos(t^{3})\mathrm {d} t.$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

[show] Hint
Try making a substitution that simplifies the exponent in the trigonometric function, and rewrite the simplified integral accordingly.

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[show] Solution

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.