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Chapter 7: Problem 8

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

Short Answer

Expert verified
Question: Evaluate the integral \(\int x \sin(2x) dx\). Answer: \(-\frac{1}{2}x\cos(2x) + \frac{1}{2}\sin(2x) + C\)

Step by step solution

01

Identify u and dv

Let \(u = x\). This makes \(dv = \sin(2x) dx\). Now, we will differentiate u and integrate dv. Step 2: Differentiate u and integrate dv
02

Calculate du and v

Differentiate \(u\) with respect to \(x\) to get \(du=dx\). Next, find the integral of \(dv\), which is \(\int\sin(2x) dx\). We need to use substitution to solve this. Let \(t=2x\). Then, \(\frac{dt}{2}=dx\). Thus, we have \(\int\sin(t)\left(\frac{dt}{2}\right)=\frac{1}{2}\int\sin(t) dt\). The integral of \(\sin(t)\) is \(-\cos(t)\). So, we have \(v=-\frac{1}{2}\cos(t)=-\frac{1}{2}\cos(2x)\). Step 3: Apply the integration by parts formula
03

Implement integration by parts

Now we can use the integration by parts formula: $$\int x \sin(2x) dx = uv - \int vdu = x\left(-\frac{1}{2}\cos(2x)\right) - \int -\frac{1}{2}\cos(2x) dx$$ Step 4: Simplify and integrate
04

Simplify and find the remaining integral

The given integral simplifies to: $$-\frac{1}{2}x\cos(2x) + \frac{1}{2}\int \cos(2x) dx$$ Now, integrate \(\cos(2x)\) using substitution. Let \(t=2x\) so \(\frac{dt}{2}=dx\), and we have \(\int\cos(t)\left(\frac{dt}{2}\right) = \frac{1}{2}\int\cos(t) dt\). The integral of \(\cos(t)\) is \(\sin(t)\). Thus, we have \(\frac{1}{2}\sin(t) = \frac{1}{2}\sin(2x)\). Step 5: Write the final answer
05

Combine the results

Adding the result from step 4 to the expression, we obtain the final answer: $$-\frac{1}{2}x\cos(2x) + \frac{1}{2}\sin(2x) + C$$ where \(C\) is the constant of integration.

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