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$Let I_{1} = \int_{a}^{π - a} xf (\sin x) dx, I_{2} = \int_{a}^{π - a} f (\sin x) dx, then$ $I_{2} is equal to$

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Correct option is C

GivenI1=∫aπ−a xf(sin⁡x)dxand I2=∫aπ−a f(sin⁡x)dx Now, I1=∫aπ−a xf(sin⁡x)dx=∫aπ−a (π−x)f[sin⁡(π−x)]dx=∫aπ−a (π−x)f(sin⁡x)dx=∫aπ−a πf(sin⁡x)dx−I1⇒ 2I1=πI2⇒ I2=2πI1

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Let I1=∫aπ−a xf(sin⁡x)dx,I2=∫aπ−a f(sin⁡x)dx, then I2 is equal to

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$Let I_{1} = \int_{a}^{π - a} xf (\sin x) dx, I_{2} = \int_{a}^{π - a} f (\sin x) dx, then$ $I_{2} is equal to$