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Using the property
∫ 0 2 a f ( x ) d x = 0 if f ( 2 a − x ) = − f ( x ) \begin{aligned} &\int_0^{2a}f(x)\,dx = 0 \\ &\text{if } f(2a - x) = -f(x) \end{aligned} ∫ 0 2 a f ( x ) d x = 0 if f ( 2 a − x ) = − f ( x ) Why?
Where the property comes from (for a function that can be integrated on [ 0 , 2 a ] [0, 2a] [ 0 , 2 a ] a a a x → 2 a − x x \to 2a - x x → 2 a − x
∫ 0 2 a f ( x ) d x = ∫ 0 a f ( x ) d x + ∫ 0 a f ( 2 a − x ) d x \begin{aligned} \int_0^{2a}f(x)\,dx &= \int_0^a f(x)\,dx \\ &\quad + \int_0^a f(2a - x)\,dx \end{aligned} ∫ 0 2 a f ( x ) d x = ∫ 0 a f ( x ) d x + ∫ 0 a f ( 2 a − x ) d x If f ( 2 a − x ) = − f ( x ) f(2a - x) = -f(x) f ( 2 a − x ) = − f ( x )
Here 2 a = 2 π 2a = 2\pi 2 a = 2 π a = π a = \pi a = π f ( x ) = cosec 7 x f(x) = \operatorname{cosec}^7x f ( x ) = cosec 7 x
f ( 2 π − x ) = cosec 7 ( 2 π − x ) = ( − cosec x ) 7 = − cosec 7 x = − f ( x ) \begin{aligned} f(2\pi - x) &= \operatorname{cosec}^7(2\pi - x) \\ &= (-\operatorname{cosec}x)^7 \\ &= -\operatorname{cosec}^7x = -f(x) \end{aligned} f ( 2 π − x ) = cosec 7 ( 2 π − x ) = ( − cosec x ) 7 = − cosec 7 x = − f ( x ) Why?
sin ( 2 π − x ) = − sin x \sin(2\pi - x) = -\sin x sin ( 2 π − x ) = − sin x cosec ( 2 π − x ) = − cosec x \operatorname{cosec}(2\pi - x) = -\operatorname{cosec}x cosec ( 2 π − x ) = − cosec x
sin ( 2 π − x ) = sin 2 π cos x − cos 2 π sin x = ( 0 ) cos x − ( 1 ) sin x = − sin x cosec ( 2 π − x ) = 1 − sin x = − cosec x \begin{aligned} &\sin(2\pi - x) \\ &= \sin 2\pi\cos x - \cos 2\pi\sin x \\ &= (0)\cos x - (1)\sin x \\ &= -\sin x \\[4pt] &\operatorname{cosec}(2\pi - x) \\ &= \frac{1}{-\sin x} = -\operatorname{cosec}x \end{aligned} sin ( 2 π − x ) = sin 2 π cos x − cos 2 π sin x = ( 0 ) cos x − ( 1 ) sin x = − sin x cosec ( 2 π − x ) = − sin x 1 = − cosec x An odd power keeps the minus sign: ( − 1 ) 7 = − 1 (-1)^7 = -1 ( − 1 ) 7 = − 1
( − cosec x ) 7 = ( − 1 ) 7 cosec 7 x = − cosec 7 x \begin{aligned} (-\operatorname{cosec}x)^7 &= (-1)^7\operatorname{cosec}^7x \\ &= -\operatorname{cosec}^7x \end{aligned} ( − cosec x ) 7 = ( − 1 ) 7 cosec 7 x = − cosec 7 x ∴ ∫ 0 2 π cosec 7 x d x = 0 \therefore\ \int_0^{2\pi}\operatorname{cosec}^7x\,dx = 0 ∴ ∫ 0 2 π cosec 7 x d x = 0 Answer: (A) 0 0 0
Strictly, cosec 7 x \operatorname{cosec}^7x cosec 7 x x = 0 x = 0 x = 0 π \pi π 2 π 2\pi 2 π 0 0 0
Where this comes from Chapter 7, Integrals : Definite integrals: the f(2a − x) property.
Question type: Limits from 0 to a a a
How to spot it: click to see Type 12. Limits from 0 to a a a f ( a − x ) f(a - x) f ( a − x )
Spot it: replacing x x x a − x a - x a − x a a a π 2 \frac{\pi}{2} 2 π
Here: Here the limits are 0 and 2 π 2\pi 2 π x x x 2 π − x 2\pi - x 2 π − x cosec 7 x \operatorname{cosec}^7x cosec 7 x − cosec 7 x -\operatorname{cosec}^7x − cosec 7 x π \pi π π \pi π 2 π 2\pi 2 π
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