Let I 1-3-6sin x - x dx - I 2-3-6sin sin x -sin x dx - I 3-3-6sin tan x -tan x d x- then- I 2- I 1- I 3- I 1- I 2- I 3- I 3- I 1- I 2D- I 3- I 2- I 1

The correct option is C I_3 lt; I_1 lt; I_2f(x) =sin x/x is a decreasing function and sin x/x gt; 0 for all x in (0, π) Since, sin x lt; x lt; ta ...

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Byju's Answer

Standard XII

Mathematics

Second Fundamental Theorem of Calculus

Let I 1=∫π/3π...

Question

Let $I_{1} = \int_{\frac{π}{3}}^{\frac{π}{6}} \frac{sin x}{x} d x, I_{2} = \int_{\frac{π}{3}}^{\frac{π}{6}} \frac{sin (sin x)}{sin x} d x, I_{3} = \int_{\frac{π}{3}}^{\frac{π}{6}} \frac{sin (tan x)}{tan x} d x$ , then

I_{1} < I_{2} < I_{3}

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I_{2} < I_{1} < I_{3}

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I_{3} < I_{1} < I_{2}

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I_{3} < I_{2} < I_{1}

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Solution

The correct option is C $I_{3} < I_{1} < I_{2}$
$f (x) = \frac{sin x}{x}$ is a decreasing function and $\frac{sin x}{x} > 0$ for all $x$ in $(0, π)$
Since, $sin x < x < tan x$
$\Rightarrow \frac{sin (sin x)}{sin x} > \frac{sin x}{x} > \frac{sin (tan x)}{tan x} for \frac{π}{6} < x < \frac{π}{3}$
$∴ I_{2} > I_{1} > I_{3}$

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Let I1=∫π6π3sinxxdx,I2=∫π6π3sin(sinx)sinxdx,I3=∫π6π3sin(tanx)tanxdx, then

The correct option is C I3<I1<I2f(x)=sinxx is a decreasing function and sinxx>0 for all x in (0,π) Since, sinx<x<tanx ⇒sin(sinx)sinx>sinxx>sin(tanx)tanx for π6<x<π3 ∴I2>I1>I3

Let $I_{1} = \int_{\frac{π}{3}}^{\frac{π}{6}} \frac{sin x}{x} d x, I_{2} = \int_{\frac{π}{3}}^{\frac{π}{6}} \frac{sin (sin x)}{sin x} d x, I_{3} = \int_{\frac{π}{3}}^{\frac{π}{6}} \frac{sin (tan x)}{tan x} d x$ , then