-2-2x5-x3cos x-sin 5 x-1 d x is equal to

The correct option is D πLet I= ∫_ π/2^π/2(x^5+x^3cos x +sin^5 x+1) dx ∫_ π/2^π/2f(x) dx=∫_ π/2^π/2(x^5+x^3cos x +sin^5 x) dx=0 nbsp; ∵ f(x) is odd function ...

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Standard XII

Mathematics

Limit

∫-π/2π/2x5+x3...

Question

$\frac{π}{2} \int - \frac{π}{2} (x^{5} + x^{3} cos x + {sin}^{5} x + 1) d x$ is equal to

0

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1

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\frac{π}{2}

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π

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Solution

The correct option is D $π$
Let
$I = \frac{π}{2} \int - \frac{π}{2} (x^{5} + x^{3} cos x + {sin}^{5} x + 1) d x$
$\frac{π}{2} \int - \frac{π}{2} f (x) d x = \frac{π}{2} \int - \frac{π}{2} (x^{5} + x^{3} cos x + {sin}^{5} x) d x = 0$ $∵ f (x) is odd function$
$∴ I = \frac{π}{2} \int - \frac{π}{2} 1 \cdot d x = \frac{π}{2} - (- \frac{π}{2}) = π$

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π2∫−π2(x5+x3cosx+sin5x+1) dx is equal to

The correct option is D πLet I=π2∫−π2(x5+x3cosx+sin5x+1) dx π2∫−π2f(x) dx=π2∫−π2(x5+x3cosx+sin5x) dx=0 ∵f(x) is odd function ∴I=π2∫−π21⋅dx=π2−(−π2)=π

$\frac{π}{2} \int - \frac{π}{2} (x^{5} + x^{3} cos x + {sin}^{5} x + 1) d x$ is equal to