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NCERT Class XII Mathematics Chapter - - Solutions

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Exercise - 7.11
By using the properties of definite integrals, evaluate the integrals in Exercises 208 to 225.
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Question : 211 of 271
Marks: +1, -0
∫0π/2cos⁡5xsin⁡5x+cos⁡5xdx\int\limits_{0}^{\pi/2}\frac{\cos^5 x}{\sin^5 x + \cos^5 x}dx
Solution:  
Let I=∫0π/2cos⁡5xsin⁡5x+cos⁡5xdxI = \int\limits_{0}^{\pi/2}\frac{\cos^5 x}{\sin^5 x + \cos^5 x}dx ...(i)
Also, I=∫0π/2cos⁡5(π/2−x)sin⁡5(π/2−x)+cos⁡5(π/2−x)dxI = \int\limits_{0}^{\pi/2}\frac{\cos^5(\pi/2 - x)}{\sin^5(\pi/2 - x) + \cos^5(\pi/2 - x)}dx [∵∫0af(x)dx=∫0af(a−x)dx]\left[\because \int\limits_{0}^{a} f(x) dx = \int\limits_{0}^{a} f(a-x) dx\right]
=∫0π/2sin⁡5xcos⁡5x+sin⁡5xdx= \int\limits_{0}^{\pi/2}\frac{\sin^5 x}{\cos^5 x + \sin^5 x}dx ...(ii)
Adding (i) and (ii), we have
2I=∫0π/2cos⁡5x+sin⁡5xcos⁡5x+sin⁡5xdx2I = \int\limits_{0}^{\pi/2}\frac{\cos^5 x + \sin^5 x}{\cos^5 x + \sin^5 x}dx =∫0π/2dx= \int\limits_{0}^{\pi/2} dx =[x]0π/2=π/2= [x]_{0}^{\pi/2} = \pi/2
Hence, I=π/4I = \pi/4
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