Let f:(0,1]→R be a continuous function such that ∫π0f(sinx)dx=2018,then∫π0xf(sinx)dx is equal to
A
1009π
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B
1008π
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C
2017π
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D
2016π
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Solution
The correct option is A1009π Let I=∫π0xf(sinx)dx⇒I=∫π0(π−x)f(sin(π−x))dx⇒I=∫π0(π−x)f(sinx)dx⇒I=π∫π0f(sinx)dx−∫π0xf(sinx)dx⇒2I=π∫π0f(sinx)dx=2018π⇒I=1009π