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NCERT Class XII Mathematics Chapter - - Solutions

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..... Miscellaneous Exercise .....
Integrate the functions in Exercises 229 to 252.
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Question : 248 of 271
Marks: +1, -0
1−x1+x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}
Solution:  
Let I=1−x1+xdxI=\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx
Put t=cos⁡t\sqrt{t}=\cos t ⇒x=cos⁡2t\Rightarrow x=\cos^2 t ⇒dx=2cos⁡t(−sin⁡t)dt\Rightarrow dx=2\cos t(-\sin t)dt
∴I=∫1−cos⁡t1+cos⁡t(−2sin⁡tcos⁡t)dt\therefore I=\int\sqrt{\frac{1-\cos t}{1+\cos t}}(-2\sin t\cos t)dt
=−4∫2sin⁡2t22cos⁡2t2sin⁡t2cos⁡t2dt=-4\int\sqrt{\frac{2\sin^2\frac{t}{2}}{2\cos^2\frac{t}{2}}}\sin\frac{t}{2}\cos\frac{t}{2}dt
=−4∫sin⁡2t2cos⁡tdt=-4\int\sin^2\frac{t}{2}\cos tdt
=−4∫1−cos⁡t2cos⁡tdt=-4\int\frac{1-\cos t}{2}\cos tdt
=−2∫(cos⁡t−cos⁡2t)dt=-2\int(\cos t-\cos^2 t)dt
=−2∫(cos⁡t−1+cos⁡2t2)dt=-2\int\left(\cos t-\frac{1+\cos 2t}{2}\right)dt
=−∫(2cos⁡t−1−cos⁡2t)dt=-\int(2\cos t-1-\cos 2t)dt
=−[2sin⁡t−t−sin⁡2t2]+C=-\left[2\sin t-t-\frac{\sin 2t}{2}\right]+C
=−[2sin⁡t−t−sin⁡tcos⁡t]+C=-\left[2\sin t-t-\sin t\cos t\right]+C
=−[21−x−cos⁡−1x−1−x⋅x+C]=-\left[2\sqrt{1-x}-\cos^{-1}\sqrt{x}-\sqrt{1-x}\cdot\sqrt{x}+C\right]
=−21−x+cos⁡−1x+x1−x+C]=-2\sqrt{1-x}+\cos^{-1}\sqrt{x}+\sqrt{x}\sqrt{1-x}+C]
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