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Chapter 17: Problem 9

Find \(\int_{1}^{2}(x+2) \sin x \mathrm{~d} x\).

Short Answer

Expert verified
Question: Evaluate the integral of \((x+2) \sin x\) with respect to \(x\) within the limits of \(1\) and \(2\). Answer: The evaluated integral is \(-4\cos(2) + \sin(2) + 3\cos(1) - \sin(1)\).

Step by step solution

01

Identify the integration by parts formula

Recall the integration by parts formula: \(\int u dv = uv-\int v du\), where \(u\) and \(dv\) are the chosen parts of the function being integrated.
02

Choose \(u\) and \(dv\)

From the function \((x+2) \sin x\), we choose: \(u = x+2\) \(dv = \sin x dx\)
03

Differentiate \(u\) and integrate \(dv\)

We differentiate \(u\) with respect to \(x\) to get its differential \(du\): \(\frac{d}{dx}(x+2) = 1\), so \(du=dx\) Next, we integrate \(dv = \sin x dx\) with respect to \(x\): \(\int \sin x dx = -\cos x + C\), so \(v=-\cos x\)
04

Apply the integration by parts formula

Now we can apply the integration by parts formula from step 1, with our values for \(u\), \(v\), and \(du\): \(\int (x+2) \sin x dx = (x+2)(-\cos x) - \int(-\cos x) dx\)
05

Calculate the remaining integral and combine terms

Integrate \(\cos x\) with respect to \(x\): \(\int (-\cos x) dx = -\sin x + C\) Substitute this result into the expression above: \(\int (x+2) \sin x dx = -(x+2)\cos x + \sin x + C\)
06

Evaluate the integral within the given limits

Now apply the given limits of integration, \(1\) and \(2\), to the integral: \(\big[-(x+2)\cos x + \sin x\big]_1^2\) = \(\big[ -(2+2)\cos(2) + \sin(2) \big] - \big[ -(1+2)\cos(1) + \sin(1) \big]\) \(=-4\cos(2) + \sin(2) + 3\cos(1) - \sin(1)\) This is the final evaluated integral within the given limits for \(\int_{1}^{2}(x+2) \sin x \mathrm{~d} x\).

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Most popular questions from this chapter

By expressing the following in partial fractions evaluate the given integral. Remember to select the correct form for the partial fractions. $$ \int \frac{13 x-4}{6 x^{2}-x-2} \mathrm{~d} x $$

If \(I_{n}=\int x^{n} \mathrm{e}^{2 x} \mathrm{~d} x\) show that \(I_{n}=\frac{x^{n} \mathrm{e}^{2 x}}{2}-\frac{n}{2} I_{n-1}\).

Use the identity \(\sin (A+B)+\sin (A-B)=2 \sin A \cos B\) to find \(\int \sin 3 x \cos 2 x \mathrm{~d} x\).

By expressing the following in partial fractions evaluate the given integral. Remember to select the correct form for the partial fractions. $$ \int \frac{1}{x^{2}-2 x-1} \mathrm{~d} x $$

Find \(\int 2 x-\mathrm{e}^{x} \mathrm{~d} x\).
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