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- Integrate by partial fractions
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- Product of Binomials with Common Term
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A binomial squared (difference) is equal to the square of the first term, minus the double product of the first by the second, plus the square of the second term. In other words: $(a-b)^2=a^2-2ab+b^2$
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$\int_{0}^{60}\left(3600-120t+t^2+\left(60-t\right)\sin\left(\sqrt{t}\right)\right)dt$
Learn how to solve problems step by step online. Integrate the function (60-t)^2+(60-t)sin(t^(1/2)) from 0 to 60. A binomial squared (difference) is equal to the square of the first term, minus the double product of the first by the second, plus the square of the second term. In other words: (a-b)^2=a^2-2ab+b^2. Multiply the single term \sin\left(\sqrt{t}\right) by each term of the polynomial \left(60-t\right). Expand the integral \int_{0}^{60}\left(3600-120t+t^2+60\sin\left(\sqrt{t}\right)-t\sin\left(\sqrt{t}\right)\right)dt into 5 integrals using the sum rule for integrals, to then solve each integral separately. The integral \int_{0}^{60}3600dt results in: 216000.
