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Official PYQ · 2015Set 1 · Q44Maxima and Minima, Mean Value Theorem, IntegrationNAT1 mark

GATE CS 2015 Set 1 Q44 — Maxima and Minima, Mean Value Theorem, Integration

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Compute the value of:

∫1π2πcos⁡(1/x)x2dx\large \int \limits_{\frac{1}{\pi}}^{\frac{2}{\pi}}\frac{\cos(1/x)}{x^{2}}dx
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Answer: -1

To evaluate the definite integral:

I=∫1π2πcos⁡(1/x)x2dxI = \int_{\frac{1}{\pi}}^{\frac{2}{\pi}} \frac{\cos(1/x)}{x^2} dx

We can use the method of substitution. Let:

u=1xu = \frac{1}{x}

Differentiating both sides with respect to xx gives:

du=−1x2dx  ⟹  1x2dx=−dudu = -\frac{1}{x^2} dx \implies \frac{1}{x^2} dx = -du

Next, we determine the new limits of integration for uu:

  • When x=1πx = \frac{1}{\pi}, u=11/π=πu = \frac{1}{1/\pi} = \pi
  • When x=2πx = \frac{2}{\pi}, u=12/π=π2u = \frac{1}{2/\pi} = \frac{\pi}{2}

Substituting these into the integral:

I=∫ππ2cos⁡(u)(−du)I = \int_{\pi}^{\frac{\pi}{2}} \cos(u) (-du)

Using the property of definite integrals, ∫abf(u)du=−∫baf(u)du\int_{a}^{b} f(u) du = -\int_{b}^{a} f(u) du, we can swap the limits to eliminate the negative sign:

I=∫π2πcos⁡(u)duI = \int_{\frac{\pi}{2}}^{\pi} \cos(u) du

Now, integrate cos⁡(u)\cos(u):

I=[sin⁡(u)]π2πI = \Big[ \sin(u) \Big]_{\frac{\pi}{2}}^{\pi} I=sin⁡(π)−sin⁡(π2)I = \sin(\pi) - \sin\left(\frac{\pi}{2}\right) I=0−1=−1I = 0 - 1 = -1

Thus, the value of the integral is -1.

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