If I1-0-2cossin xdx - I2-0-2sincos xdx and I3-0-2cos xdx- then

The correct option is A I_1>I_3>I_2 I_1=∫ _0^Π/2 cos (sin x) dx; ∫ _0^Π/2 sin (cos x) dx. I_3=∫ _0^Π/2 cos x dx. So, I_1 gt; I_3 gt; ...

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Geometric Interpretation of Def.Int as Limit of Sum

If I1=∫0π/2...

Question

If $I_{1} = \int_{0}^{π / 2} cos (sin x) d x$ ; $I_{2} = \int_{0}^{π / 2} sin (cos x) d x$ and $I_{3} = \int_{0}^{π / 2} cos x d x$ , then

I_{1} > I_{3} > I_{2}

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I_{3} > I_{1} > I_{2}

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I_{1} > I_{2} > I_{3}

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I_{3} > I_{2} > I_{1}

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Solution

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I_{1} = π / 2 \int 0 cos (sin x) d x; I_{2} = π / 2 \int 0 sin (cos x) d x

I_{3} = π / 2 \int 0 cos x d x,

I_{1} = \int_{0}^{\frac{π}{4}} x^{2008} (t a n x)^{2008} d x, I_{2} = \int_{0}^{\frac{π}{4}} x^{2009} (t a n x)^{2009} d x a n d I_{3} = \int_{0}^{\frac{π}{4}} x^{2010} (t a n x)^{2010} d x

I_{1} = \int_{0}^{\frac{π}{2}} c o s (π s i n^{2} x) d x, I_{2} = \int_{0}^{\frac{π}{2}} c o s (2 π s i n^{2} x) d x a n d I_{3} = \int_{0}^{\frac{π}{2}} c o s (π s i n x) d x,

I_{1} = \int_{0}^{π / 2} {sin}^{4} x d x

I_{2} = \int_{0}^{π / 2} {cos}^{6} x d x

I_{3} = \int_{0}^{π / 2} {sin}^{8} x d x

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I_{1}, I_{2}, I_{3}, I_{4}

I_{1} = \int_{0}^{\frac{π}{4}} e^{x^{2}} d x, I_{2} = \int_{0}^{\frac{π}{4}} e^{x} d x, I_{3} = \int_{0}^{\frac{π}{4}} e^{x^{2}} cos x d x, I_{4} = \int_{0}^{\frac{π}{4}} e^{x^{2}} sin x d x

I_{2} > I_{1} > I_{3} > I_{4} .

x (0, 1), x > x^{2}

sin x > cos x .

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