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Compute the value of:

$$\large \int_{\frac{1}{\pi}}^{\frac{2}{\pi}}\frac{\cos(1/x)}{x^{2}}dx$$
asked in Calculus by Veteran (29.1k points)  
edited by | 743 views

2 Answers

+25 votes
Best answer
For the integrand $\frac{\cos(1/x)}{x^2}$, substitute $u = \frac1 x$ and $\def\d{\,\mathrm{d}} \d u = -\frac1{x^2}\d x$.

This gives a new lower bound $u = \frac1{1/\pi} = \pi$ and upper bound $u = \frac1{2/\pi} = \frac{\pi}{2}$. Now, our integral becomes:

$$\begin{align}
I &= - \int\limits_\pi^{\pi/2} \cos(u) \d u\\[1em]
&= \int\limits_{\pi/2}^\pi \cos(u)\d u
\end{align}$$

Since the antiderivative of $\cos(u)$ is $\sin(u)$, applying the fundamental theorem of calculus, we get:

$$\begin{align}
I &= \sin(u)\biggr |_{\pi/2}^\pi\\[1em]
&= \sin(\pi) - \sin \left ( \frac\pi 2\right )\\[1em]
&= 0 - 1\\[1em]
I &= -1
\end{align}$$
answered by Active (1.5k points)  
selected by
+6 votes
$\begin{align*} \large \int_{\frac{1}{\pi}}^{\frac{2}{\pi}}\frac{cos(1/x)}{x^{2}}dx \end{align*}$

let, $\frac{1}{x} = t$ then, $\frac{-1}{x^2}dx = dt$

$\begin{align*} \int_{\frac{1}{\pi}}^{\frac{2}{\pi}}\frac{cos(1/x)}{x^{2}}dx &=-\int_{\pi}^{\frac{\pi}{2}}cos(t)\ dt\\ &= -\Big( \sin t \Big)_{\pi}^{\pi/2}\\ &= -\Big( 1-0 \Big)\\ &= -1 \end{align*}$
answered by Veteran (27.5k points)  
But in exam they give incorrect on -1 as answer. Please improve that! I'm talking about GO test for this paper!


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