GATE CSE 2015 | Set 1 | Question: 44

Question

Compute the value of:

$$ \large \int \limits_{\frac{1}{\pi}}^{\frac{2}{\pi}}\frac{\cos(1/x)}{x^{2}}dx$$

Comments

take sin(1/x) = t and differentiate both the sides w.r.t 't'

cos(1/x)*(-1/x²)*dx/dt = 1

so, -dt = (cos(1/x)/x²)dx   (same as what is asked in question)

so whole thing can be written in terms of -dt and will integrate to -t 

i.e sin(1/x) area from 1/π to 2/π i.e -(sin(π/2)-sin(π)) = -1

P.S. Its my first answer here don't know how to use special character.

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@shadymeee you can take -1/x as t

 substitute t=-1/𝑥 and dt= d𝑥/𝑥^2.

new lower bound t=-𝜋 and upper bound t=-𝜋/2

I= sin⁡(-𝜋/2)−sin⁡(-𝜋)

I= -sin⁡(𝜋/2)+sin⁡(𝜋)
I=-1+0  = -1

 

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Commenting so that I won't forget here how dumb I was, that could not change limits after taking 1/x = t

Answer 1

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:

$I= \sin(u)\;\mid _{\pi/2}^\pi$
$\;\;= \sin(\pi) - \sin \left ( \frac\pi 2\right )$
$\;\;= 0 - 1$
$\;\; = {-1}$

Answer 2

$\begin{align*} \large \int_{\frac{1}{\pi}}^{\frac{2}{\pi}}\frac{cos(1/x)}{x^{2}}dx \end{align*}$

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

$\begin{align*} \int_{\frac{1}{\pi}}^{\frac{2}{\pi}}\frac{cos(1/x)}{x^{2}}dx &=-\int_{\pi}^{\frac{\pi}{2}}cos(t)\ dt\\ &= -\Big( \sin t \Big)_{\pi}^{\pi/2}\\ &= -\Big( 1-0 \Big)\\ &= -1 \end{align*}$

Comments

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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$\int_{b}^{a}f(x)dx=-\int_{a}^{b}f(x)dx$

Answer 3

Here is the solution : 

Answer 4

Ans (-1)