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- Integrate by partial fractions
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- Product of Binomials with Common Term
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Rewrite the integrand $\left(x^2-2x+3\right)\left(\cos\left(3x\right)+\sin\left(3x\right)\right)$ in expanded form
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$\int\left(x^2\cos\left(3x\right)+x^2\sin\left(3x\right)-2x\cos\left(3x\right)-2x\sin\left(3x\right)+3\cos\left(3x\right)+3\sin\left(3x\right)\right)dx$
Learn how to solve problems step by step online. Find the integral int((x^2-2x+3)(cos(3x)+sin(3x)))dx. Rewrite the integrand \left(x^2-2x+3\right)\left(\cos\left(3x\right)+\sin\left(3x\right)\right) in expanded form. Expand the integral \int\left(x^2\cos\left(3x\right)+x^2\sin\left(3x\right)-2x\cos\left(3x\right)-2x\sin\left(3x\right)+3\cos\left(3x\right)+3\sin\left(3x\right)\right)dx into 6 integrals using the sum rule for integrals, to then solve each integral separately. The integral \int x^2\cos\left(3x\right)dx results in: \frac{1}{3}x^2\sin\left(3x\right)+\frac{2}{9}x\cos\left(3x\right)-\frac{2}{27}\sin\left(3x\right). The integral \int x^2\sin\left(3x\right)dx results in: -\frac{1}{3}x^2\cos\left(3x\right)+\frac{2}{9}x\sin\left(3x\right)+\frac{2}{27}\cos\left(3x\right).
