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Question

02πsinx+sinxdx=


A

0

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B

4

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C

8

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D

1

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Solution

The correct option is B

4


Explanation for correct answer:

Integrating 02πsinx+sinxdx:

Let I=02πsinx+sinxdx

Using the property of mod we can write the above integral in a simplified way shown below

I=02πsinxdx+02πsinxdx=-cosx02π+0πsinxdx+π2π-sinxdxsinxdx=-cosx=-cos2π--cos0+-cosx0π--cosxπ2π=-cos2π+cos0+-cosπ+cos0--cos2π+cosπ=-cos2π+cos0-cosπ+cos0+cos2π-cosπ=2cos0-2cosπ=2×1-2×(-1)I=4

Therefore, the value of 02πsinx+sinxdx= 4

Hence, option (B) is the correct answer.


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