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    • https://phys.libretexts.org/Courses/Georgia_State_University/GSU-TM-Physics_I_(2211)/02%3A_Vectors_and_Math_Review_Topics/2.07%3A_Math_Review_of_Other_Topics/2.7.22%3A_Integrals
      \[ \begin{aligned} \int_{0}^{\pi} x^{2} \cos x d x &=-\left.\dfrac{d^{2} I(a)}{d a^{2}}\right|_{a=1} \\ &=-\left.\left.\dfrac{d^{2}}{d a^{2}}\left(\dfrac{\sin a \pi}{a}\right)\right|_{a=1}\right|_{a=1...\[ \begin{aligned} \int_{0}^{\pi} x^{2} \cos x d x &=-\left.\dfrac{d^{2} I(a)}{d a^{2}}\right|_{a=1} \\ &=-\left.\left.\dfrac{d^{2}}{d a^{2}}\left(\dfrac{\sin a \pi}{a}\right)\right|_{a=1}\right|_{a=1} \\ &\left.=-\left.\dfrac{d}{d a}\left(\dfrac{a \pi \cos a \pi-\sin a \pi}{a^{2}}\right)\right|_{a^{3}}\right)\left.\right|_{a=1} \\ &=-\left(\dfrac{a^{2} \pi^{2} \sin a \pi+2 a \pi \cos a \pi-2 \sin a \pi}{-2 \pi .}\right. \end{aligned} \label{A.75} \]
    • https://phys.libretexts.org/Courses/Georgia_State_University/GSU-TM-Physics_II_(2212)/02%3A_Math_Review/2.10%3A_Anti_derivatives_and_integrals/2.10.01%3A_Integrals
      Integration is typically a bit harder. Imagine being given the last result in Equation 8.4.2 and having to figure out what was differentiated in order to get the given function.

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