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

A
$\frac{π}{2} I_{1}$
B
${πI}_{1}$
C
$\frac{2}{π} I_{1}$
D
$2 I_{1}$

Solution:

$\begin{aligned} {GivenI}_{1} = \int_{a}^{π - a} xf (\sin x) dx \\ and I_{2} = \int_{a}^{π - a} f (\sin x) dx \end{aligned}$

$\begin{array}{l} Now, I_{1} = \int_{a}^{π - a} xf (\sin x) dx \\ = \int_{a}^{π - a} (π - x) f [\sin (π - x)] dx \\ = \int_{a}^{π - a} (π - x) f (\sin x) dx \\ = \int_{a}^{π - a} πf (\sin x) dx - I_{1} \\ \Rightarrow 2 I_{1} = {πI}_{2} \\ \Rightarrow I_{2} = \frac{2}{π} I_{1} \end{array}$

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

Solution:

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