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Question: Let \(I _ { 2 }\) is equal to...

Let I2I _ { 2 } is equal to

A

π2I1\frac { \pi } { 2 } I _ { 1 }

B

πI1\pi I _ { 1 }

C

2πI1\frac { 2 } { \pi } I _ { 1 }

D

2I12 I _ { 1 }

Answer

2πI1\frac { 2 } { \pi } I _ { 1 }

Explanation

Solution

I1=∫aπ−axf(sin⁡x)dxI _ { 1 } = \int _ { a } ^ { \pi - a } x f ( \sin x ) d x =∫aπ−a(π−x)f(sin⁡(π−x))dx= \int _ { a } ^ { \pi - a } ( \pi - x ) f ( \sin ( \pi - x ) ) d x,

[∵∫abf(x)dx=∫abf(a+b−x)dx]\left[ \because \int _ { a } ^ { b } f ( x ) d x = \int _ { a } ^ { b } f ( a + b - x ) d x \right]

=∫aπ−a(π−x)f(sin⁡x)dx= \int _ { a } ^ { \pi - a } ( \pi - x ) f ( \sin x ) d x =∫aπ−aπf(sin⁡x)dx−I1= \int _ { a } ^ { \pi - a } \pi f ( \sin x ) d x - I _ { 1 }

⇒2I1=πI2⇒I2=2πI1\Rightarrow 2 I _ { 1 } = \pi I _ { 2 } \Rightarrow I _ { 2 } = \frac { 2 } { \pi } I _ { 1 } .