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  • The value of <img alt="\int_{0}^{2x}\frac{x\sin ^{8}x}{\sin ^{8}x+\cos ^{8}x}dx" src="https://learn.careers360.com/latex-image/?%5Cint_%7B0%7D%5E%7B2x%7D%5Cfrac%7Bx%5Csin%20%5E%7B8%7Dx%7D%7B%5Csin%20%5E%7B8%7Dx&plus;%5Ccos%20%5E%7B8%7Dx%7Ddx"

The value of \int_{0}^{2x}\frac{x\sin ^{8}x}{\sin ^{8}x+\cos ^{8}x}dx is equal to : 
Option: 1 2\pi
 
Option: 2 4\pi
 
Option: 3 2\pi ^{2}
 
Option: 4 \pi ^{2}
 
 

Answers (1)

best_answer

 

 

Properties of the Definite Integral (Part 2) - King's Property -

Property 4 (King's Property)

This is one of the most important properties of definite integration.

\\\mathbf{\int_{a}^{b}f(x)\;dx=\int_{a}^{b}f(a+b-x)\;dx}

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Application of Periodic Properties in Definite Integration -

Property 9

If f(x) is a periodic function with period T, then the area under f(x) for n periods would be n times the area under f(x) for one period, i.e.

\mathbf{\int_{0}^{n T} f(x) d x=n \int_{0}^{T} f(x) d x}

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\\I=\int _0^{2\pi }\frac{x\sin ^8x\:}{\sin ^8x+\cos ^8x\:\:}dx\\I=\int _0^{2\pi }\frac{(2\pi-x)\sin ^8(2\pi-x)\:}{\sin ^8(2\pi-x)+\cos ^8(2\pi-x)\:\:}dx\\2I=\int _0^{2\pi }\frac{2\pi \sin ^8x\:}{\sin ^8x+\cos ^8x\:\:}dx\\I=\int _0^{2\pi }\frac{\pi \sin ^8x\:}{\sin ^8x+\cos ^8x\:\:}dx

\\I=\int _0^{2\pi }\frac{\pi \sin ^8x\:}{\sin ^8x+\cos ^8x\:\:}dx\\I=4\int _0^{\pi/2 }\frac{\pi \sin ^8x\:}{\sin ^8x+\cos ^8x\:\:}dx\\I=4\int _0^{\pi/2 }\frac{\pi \sin ^8(\pi/2-x)\:}{\sin ^8(\pi/2-x)+\cos ^8(\pi/2-x)\:\:}dx\\I=2\int_{0}^{\pi/2}\pi dx\\I=\pi^2

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