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Compute the value of:

$$\large \int_{\frac{1}{\pi}}^{\frac{2}{\pi}}\frac{\cos(1/x)}{x^{2}}dx$$
in Calculus by Boss (30.8k points)
edited by | 2.4k views

2 Answers

+38 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:

$I= - \int\limits_\pi^{\pi/2} \cos(u) \d u$

$\;\;= \int\limits_{\pi/2}^\pi \cos(u)\d u$

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

$I= \sin(u)\;\mid _{\pi/2}^\pi$
$\;\;= \sin(\pi) - \sin \left ( \frac\pi 2\right )$
$\;\;= 0 - 1$
$\;\; = {-1}$
by Active (1.6k points)
edited by
+13 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, $\frac{-1}{x^2}dx = dt$

$\begin{align*} \int_{\frac{1}{\pi}}^{\frac{2}{\pi}}\frac{cos(1/x)}{x^{2}}dx &=-\int_{\pi}^{\frac{\pi}{2}}cos(t)\ dt\\ &= -\Big( \sin t \Big)_{\pi}^{\pi/2}\\ &= -\Big( 1-0 \Big)\\ &= -1 \end{align*}$
by Boss (30.8k points)
+3
But in exam they give incorrect on -1 as answer. Please improve that! I'm talking about GO test for this paper!
+1
$\int_{b}^{a}f(x)dx=-\int_{a}^{b}f(x)dx$
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