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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 : 218 of 271
Marks: +1, -0
∫0πxdx1+sin⁡x\int\limits_{0}^{\pi} \frac{xdx}{1+\sin x}
Solution:  
Let I=∫0πxdx1+sin⁡xI=\int\limits_{0}^{\pi} \frac{xdx}{1+\sin x} ...(i)
I=∫0ππ−x1+sin⁡(π−x)dxI=\int\limits_{0}^{\pi} \frac{\pi - x}{1+\sin(\pi - 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]
=∫0ππ−x1+sin⁡xdx= \int\limits_{0}^{\pi} \frac{\pi - x}{1+\sin x} dx...(ii)
Adding (i) and (ii), we get
2I=∫0πx+π−x1+sin⁡xdx2I = \int\limits_{0}^{\pi} \frac{x+\pi - x}{1+\sin x} dx
=π∫0π11+sin⁡xdx= \pi \int\limits_{0}^{\pi} \frac{1}{1+\sin x} dx
=π∫0π1−sin⁡x1−sin⁡2xdx= \pi \int\limits_{0}^{\pi} \frac{1-\sin x}{1-\sin^2 x} dx
=π∫0π1−sin⁡xcos⁡2xdx= \pi \int\limits_{0}^{\pi} \frac{1-\sin x}{\cos^2 x} dx
=π∫0π(sec⁡2x−tan⁡xsec⁡x)dx= \pi \int\limits_{0}^{\pi} (\sec^2 x - \tan x \sec x) dx
=π[tan⁡x−sec⁡x]0π= \pi \left[ \tan x - \sec x \right]_{0}^{\pi}
=π[(tan⁡π−sec⁡π)−(tan⁡0−sec⁡0)]= \pi \left[ (\tan \pi - \sec \pi) - (\tan 0 - \sec 0) \right]
=π[(0−(−1))−(0−1)]=2π= \pi \left[ (0 - (-1)) - (0 - 1) \right] = 2\pi
⇒I=π\Rightarrow I = \pi
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