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Chapter 7: Problem 29

Evaluate the following integrals. $$\int x^{2} \sin 2 x d x$$

Short Answer

Expert verified
Answer: The result of the integral is \(\int x^{2} \sin 2x \, dx = -\frac{1}{2}x^{2}\cos 2x + \frac{1}{2}x \sin 2x - \frac{1}{4} \cos 2x + C\), where \(C\) is the constant of integration.

Step by step solution

01

Choose u and dv for the first application of integration by parts

Choose \(u = x^{2}\) and \(dv = \sin 2x \, dx\).
02

Calculate du and v

Compute \(du\) by differentiating \(u\) with respect to \(x\) and compute \(v\) by integrating \(dv\) with respect to \(x\): $$du = 2x \, dx$$ $$v = -\frac{1}{2} \cos 2x$$
03

Apply integration by parts formula

Substitute the values of \(u, v, du\) and \(dv\) into the integration by parts formula: $$\int x^{2} \sin 2x \, dx = -\frac{1}{2}x^{2}\cos 2x - \int-\frac{1}{2}\cos 2x (2x) \, dx$$ Simplify the expression: $$\int x^{2} \sin 2x \, dx = -\frac{1}{2}x^{2}\cos 2x + x \int\cos 2x \, dx$$
04

Choose u and dv for the second application of integration by parts

For the remaining integral, choose \(u = x\) and \(dv = \cos 2x \, dx\).
05

Calculate du and v again

Compute \(du\) by differentiating \(u\) with respect to \(x\) and compute \(v\) by integrating \(dv\) with respect to \(x\): $$du = dx$$ $$v = \frac{1}{2} \sin 2x$$
06

Apply integration by parts formula again

Substitute the new values of \(u, v, du\) and \(dv\) into the integration by parts formula for the remaining integral: $$x \int \cos 2x \, dx = \frac{1}{2}x \sin 2x - \int \frac{1}{2} \sin 2x \, dx$$ Integrate \(\frac{1}{2}\sin 2x\) with respect to \(x\): $$x \int \cos 2x \, dx = \frac{1}{2}x \sin 2x - \frac{1}{4} \cos 2x$$
07

Combine results from second integration by parts

Combine the result from Step 3 with the result from Step 6: $$\int x^{2} \sin 2x \, dx = -\frac{1}{2}x^{2}\cos 2x + \left(\frac{1}{2}x \sin 2x - \frac{1}{4} \cos 2x\right) + C$$ where \(C\) is the constant of integration.
08

Final answer

The result of the integral is: $$\int x^{2} \sin 2x \, dx = -\frac{1}{2}x^{2}\cos 2x + \frac{1}{2}x \sin 2x - \frac{1}{4} \cos 2x + C$$

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