Mathematics

# The value of $\displaystyle\int^{2\pi}_{0}\dfrac{x\sin^8x}{\sin^8x+\cos^8x}dx$ is equal to?

$\pi^2$

##### SOLUTION
Let $I = \displaystyle \int_0^{2 \pi} \dfrac{x \sin^8 x}{\sin^8 x + \cos^8 x}$ ___(1)

$I = \displaystyle \int_0^{2 \pi} \dfrac{(2\pi - x) \sin^8 (2\pi - x)}{\sin^8 (2 \pi - x) + \cos^8 (2 \pi - x)}$

$I = \displaystyle \int_0^{2 \pi} \dfrac{(2\pi - x) \sin^8 x}{\sin^8 x + \cos^8 x}$ ___(2)

Add (1) + (2) we get,

$2I = \displaystyle \int_0^{2 \pi} \dfrac{2 \pi \sin^8 x}{\sin^8 x + \cos^8 x}$

$2I = 2 \times 2 \pi \displaystyle \int_0^{\pi} \dfrac{\sin^8 x}{\sin^8 x + \cos^8 x}$

$I = 2 \pi \displaystyle \int_0^{\pi} \dfrac{\sin^8 x}{\sin^8 x + \cos^8 x} dx$

$I = 4 \pi \displaystyle \int_0^{\pi/2} \dfrac{\sin^8 x}{\sin^8 x + \cos^8 x} dx$ __(3)

$I = 4\pi \displaystyle \int_0^{\pi/2} \dfrac{\sin^8 \left(\dfrac{\pi}{2} - x \right)}{\sin^8 \left(\dfrac{\pi}{2} - x \right) + \cos^8 \left(\dfrac{\pi}{2} - x \right)}$

$I = 4 \pi \displaystyle \int_0^{\pi/2} \dfrac{\cos^8 (x)}{\sin^8 x + \cos^8 x}$ __(4)

add (3) + (4) we get

$I = 2 \pi \displaystyle \int_0^{\pi/2} \dfrac{\sin^8 x + \cos^8 x}{\sin^8 x + \cos^8 x} dx = 2\pi \displaystyle \int_0^{\pi/2} 1 dx = 2 \pi \times \dfrac{\pi}{2} = \pi^2$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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