Evaluate the Integral integral of x^2cos(3x) with respect to x

Problem

(∫_^)(x2*cos(3*x)*d(x))

Solution

1. Identify the method of integration by parts, which states (∫_^)(u*d(v))=u*v−(∫_^)(v*d(u)) Let u=x2 and d(v)=cos(3*x)*d(x)

2. Differentiate u to find d(u)=2*x*d(x) and integrate d(v) to find v=1/3*sin(3*x)

3. Apply the formula for the first time:

(∫_^)(x2*cos(3*x)*d(x))=1/3*x2*sin(3*x)−(∫_^)(2/3*x*sin(3*x)*d(x))

4. Apply integration by parts again to the new integral (∫_^)(2/3*x*sin(3*x)*d(x)) Let u=2/3*x and d(v)=sin(3*x)*d(x)

5. Differentiate u to find d(u)=2/3*d(x) and integrate d(v) to find v=−1/3*cos(3*x)

6. Substitute these values back into the expression:

(∫_^)(2/3*x*sin(3*x)*d(x))=−2/9*x*cos(3*x)−(∫_^)(−2/9*cos(3*x)*d(x))

7. Simplify the double negative and integrate the final term:

(∫_^)(2/3*x*sin(3*x)*d(x))=−2/9*x*cos(3*x)+2/27*sin(3*x)

8. Combine all parts into the original equation and add the constant of integration C

(∫_^)(x2*cos(3*x)*d(x))=1/3*x2*sin(3*x)−(−2/9*x*cos(3*x)+2/27*sin(3*x))+C

9. Distribute the negative sign to reach the final form.

Final Answer

(∫_^)(x2*cos(3*x)*d(x))=1/3*x2*sin(3*x)+2/9*x*cos(3*x)−2/27*sin(3*x)+C

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