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Evaluate-

$$\int_\pi^{2\pi}\frac{(x^2+2)(\cos x)}{x^3}dx$$

Separating the function as $\frac{\cos x}{x}+\frac{2\cos x}{x^3}$ and evaluating the indefinite integral and putting the limits does give an answer.

But is there any other more elegant way out?

Thanks for any help!

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Use integration by parts twice on $\int\ \frac{1}{x}\cdot \cos\ x\ dx$ so that we can rule out $\frac{2\cos\ x}{x^3}$

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The indefinite integral,

$I = \int(\frac{\cos x}{x}+\frac{2\cos x }{x^3})dx = \int(\frac{\cos x}{x})dx+\int(\frac{2\cos x}{x^3})dx$

$I_1 = \int(\frac{\cos x}{x})dx$

Let $u= \frac{1}{x}, dv = \cos x dx$

$du = -\frac{1}{x^2}dx, v = \sin x$

$I_1 = uv - \int vdu = \sin x\cdot\frac{1}{x}-\int{-\frac{1}{x^2}\cdot\sin xdx} +c$

$$I_1 = \frac{\sin x}{x} + \int\frac{1}{x^2}\cdot\sin xdx+c$$

$I_2 = \int(\frac{2\cos x}{x^3})dx$

Let $u = \cos x , dv = \frac{2}{x^3}dx$

$du = -\sin x dx, v = -\frac{1}{x^2}$

$I_2 = uv - \int vdu = -\frac{1}{x^2}\cos x - \int\frac{1}{x^2}\cdot\sin x dx+c'$

$$I_2 = -\frac{\cos x}{x^2} - \int\frac{1}{x^2}\cdot\sin xdx+c'$$

$$I=I_1+I_2 = \frac{\sin x}{x} -\frac{\cos x}{x^2} +C$$

Now substitute the limits and find the answer.

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Using chain rule for the first integral,

$$\int \frac{1}{x}\cos x\,dx = \frac{1}{x}\sin x + \int \frac{1}{x^2}\sin x\, dx$$

Use the chain rule again for the integral on RHS.

$$\int \frac{1}{x}\cos x\,dx = \frac{1}{x}\sin x + \frac{1}{x^2}(-\cos x) - \int \frac{-2}{x^3}(-\cos x) dx$$

With some manipulation,

$$\int \frac{1}{x}\cos x\,dx + \int \frac{2}{x^3}\cos x\, dx = \frac{\sin x}{x} - \frac{\cos x}{x^2}$$

Edit
I noticed that similar answers were posted while I was typing my answer.

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