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Evaluate the integral.Answer:
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Q. Evaluate the integral.Answer:
- Integration by Parts: Let's use integration by parts to solve the integral of . The integration by parts formula is . We will let and . Then we need to find and .
- Find : First, we differentiate to find . The derivative of with respect to is , so .
- Find : Next, we integrate to find . The integral of with respect to is , so .
- Apply Integration by Parts: Now we apply the integration by parts formula: . Substituting the values we have:$\int \(4\)x^\(2\)\sin(\(3\)x)\,dx = x^\(2\)\left(-\frac{\(4\)}{\(3\)}\cos(\(3\)x)\right) - \int\left(-\frac{\(4\)}{\(3\)}\cos(\(3\)x)\right)(\(2\)x)\,dx.
- Simplify Expression: Simplify the expression: \(\int 4x^2\sin(3x)\,dx = -\frac{4}{3}x^2\cos(3x) + \left(\frac{8}{3}\right)\int x \cos(3x)\,dx\). Now we need to integrate \(\left(\frac{8}{3}\right)\int x \cos(3x)\,dx\), which again requires integration by parts.
- New Integration by Parts: For the new integration by parts, let \(u = x\) and \(dv = \left(\frac{8}{3}\right)\cos(3x)dx\). Then \(du = dx\) and \(v = \left(\frac{8}{3}\right)\left(\frac{1}{3}\right)\sin(3x) = \left(\frac{8}{9}\right)\sin(3x)\).
- Apply Integration by Parts: Apply the integration by parts formula to the new integral: \(\frac{8}{3}\int x \cos(3x)\,dx = \frac{8}{9}x\sin(3x) - \frac{8}{9}\int \sin(3x)\,dx\).
- Integrate \(\sin(3x)\): Now we integrate \((8/9)\int\sin(3x)\,dx\). The integral of \(\sin(3x)\) with respect to \(x\) is \(-1/3\cos(3x)\), so we have:\(\newline\)\((8/9)\int\sin(3x)\,dx = -(8/9)(1/3)\cos(3x) = -(8/27)\cos(3x)\).
- Substitute Result: Substitute the result back into our expression:\(\newline\)\(\int 4x^2\sin(3x)\,dx = -\frac{4}{3}x^2\cos(3x) + \frac{8}{9}x\sin(3x) - \frac{8}{27}\cos(3x) + C\), where \(C\) is the constant of integration.
