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Compute the value of:

$$\large \int_{\frac{1}{\pi}}^{\frac{2}{\pi}}\frac{\cos(1/x)}{x^{2}}dx$$
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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) 8 12 20
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\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 (29.3k points) 75 220 379
But in exam they give incorrect on -1 as answer. Please improve that! I'm talking about GO test for this paper!