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Question

11, sin2x dx

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Solution

The integral is given as,

y= π 2 π 2 sin 2 x dx

We have to calculate the integral of y.

Consider g( x )= sin 2 x,

g( x )= sin 2 x g( x )= sin 2 ( x ) g( x )= ( sinx ) 2 g( x )= sin 2 x

It can be observed that g( x )=g( x ). So, sin 2 x is an even function.

Use the property of even function b b g( x )dx =2 0 b g( x )dx to solve the integral.

y=2 0 π 2 sin 2 x dx =2 0 π 2 ( 1cos2x 2 ) dx = 0 π 2 ( 1cos2x ) dx

Simplify further,

y= [ x sin2x 2 ] 0 π 2 = π 2 sin2( π 2 ) 2 0+ sin2( 0 ) 2 = π 2

Thus, the value of integral is π 2 .


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