The correct option is A 1 ∫_0^(π/2) sin2x #10;dx =[( cos2x/2)]_0^(π/2) = (1/2)[cos(2×(π/2)) cos0 #10;] = (1/2)×( 1 1)=1 #10; #10 ...

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Complex Numbers

∫0π2sin2x dx ...

Question

$\frac{π}{2} \int 0 sin 2 x d x$ is-

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Solution

The correct option is A $1$
$\int_{0}^{(\frac{π}{2})} s i n 2 x d x = {[(\frac{- c o s 2 x}{2})]}_{0}^{(\frac{π}{2})} = - (\frac{1}{2}) [c o s (2 \times (\frac{π}{2})) - c o s 0] = - (\frac{1}{2}) \times (- 1 - 1) = 1$

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