The value of -0-2cos2018 -sin2x-cos1009 -sin xdx is

∫_0^π/2[cos(2018 πsin^2x)+cos(1009 πsin x)]dx Let I_1=∫_0^π/2cos(2018 πsin^2x)dx and I_2 = ∫_0^π/2cos(1009 πsin x)dx Now, I_1=∫_0^π/2cos(1009π(1 cos 2x))dx ⇒ ...

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Byju's Answer

Standard XII

Mathematics

Property 5

The value of ...

Question

The value of $π / 2 \int 0 (cos (2018 π {sin}^{2} x) + cos (1009 π sin x)) d x$ is

0

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\frac{2}{π}

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\frac{π}{2}

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Solution

The correct option is A $0$
$π / 2 \int 0 [cos (2018 π {sin}^{2} x) + cos (1009 π sin x)] d x$
Let $I_{1} = π / 2 \int 0 cos (2018 π {sin}^{2} x) d x$
and $I_{2} = π / 2 \int 0 cos (1009 π sin x) d x$
Now, $I_{1} = π / 2 \int 0 cos (1009 π (1 - cos 2 x)) d x$
$\Rightarrow I_{1} = - π / 2 \int 0 cos (1009 π cos 2 x) d x \Rightarrow I_{1} = - 2 π / 4 \int 0 cos (1009 π cos 2 x) d x$
Assuming $2 x = t \Rightarrow 2 d x = d t$
$\Rightarrow I_{1} = - π / 2 \int 0 cos (1009 π cos t) d t$
Putting $t \to \frac{π}{2} - t,$ we get
$I_{1} = - I_{2} ∴ I_{1} + I_{2} = 0$

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The value of π/2∫0(cos(2018πsin2x)+cos(1009πsinx))dx is

The value of $π / 2 \int 0 (cos (2018 π {sin}^{2} x) + cos (1009 π sin x)) d x$ is