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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 : 220 of 271
Marks: +1, -0
∫02πcos⁡5x dx\int\limits_{0}^{2\pi}\cos ^5x\,dx
Solution:  
Let I=∫02πcos⁡5x dxI=\int\limits_{0}^{2\pi}\cos ^5x\,dx
Let f(x)=cos⁡5xf(x) = \cos^5x
Now we have
f(2π−x)=(cos⁡(2π−x))5f(2\pi - x) = (\cos(2\pi - x))^5 =(cos⁡x)5=cos⁡5x=f(x)= (\cos x)^5 = \cos^5x = f(x)
⇒I=2∫0πcos⁡5x dx      [∵∫02af(x)dx=2∫0af(x)dx, if f(2a−x)=f(x)]\Rightarrow I=2\int\limits_{0}^{\pi}\cos ^5x\,dx \;\;\;\left[\because \int\limits_{0}^{2a}f(x)dx=2\int\limits_{0}^{a}f(x)dx,\text{ if }f(2a-x)=f(x)\right]
Again, we have
f(π−x)=(cos⁡(π−x))5f(\pi - x) = (\cos(\pi - x))^5 =−cos⁡5x=−f(x)= -\cos5x = -f(x)
⇒2∫0πcos⁡5x dx=0\Rightarrow 2\int\limits_{0}^{\pi}\cos ^5x\,dx = 0
Hence,∫02πcos⁡5x dx=2∫0πcos⁡5x dx\int\limits_{0}^{2\pi}\cos ^5x\,dx = 2\int\limits_{0}^{\pi}\cos ^5x\,dx =2×0=0=2\times 0=0
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