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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 : 247 of 271
Marks: +1, -0
sin⁡−1x−cos⁡−1xsin⁡−1x+cos⁡−1x\frac{\sin^{-1}\sqrt{x} - \cos^{-1}\sqrt{x}}{\sin^{-1}\sqrt{x} + \cos^{-1}\sqrt{x}} , x∈[0,1]x \in [0,1]
Solution:  
Let I=∫sin⁡−1x−cos⁡−1xsin⁡−1x+cos⁡−1xdxI = \int \frac{\sin^{-1}\sqrt{x} - \cos^{-1}\sqrt{x}}{\sin^{-1}\sqrt{x} + \cos^{-1}\sqrt{x}} dx
=∫sin⁡−1x−(π2−sin⁡−1x)π2dx= \int \frac{\sin^{-1}\sqrt{x} - \left(\frac{\pi}{2} - \sin^{-1}\sqrt{x}\right)}{\frac{\pi}{2}} dx
[∵sin⁡−1x+cos⁡−1x=π2]\left[\because \sin^{-1}\sqrt{x} + \cos^{-1}\sqrt{x} = \frac{\pi}{2}\right]
=2π∫[2sin⁡−1x−π2]dx= \frac{2}{\pi} \int \left[2\sin^{-1}\sqrt{x} - \frac{\pi}{2}\right] dx
=4π∫sin⁡−1xdx−∫1dx= \frac{4}{\pi} \int \sin^{-1}\sqrt{x} dx - \int 1 dx...(i)
Let I1=∫sin⁡−1xdxI_1 = \int \sin^{-1}\sqrt{x} dx
Put x=t⇒x=t2\sqrt{x} = t \Rightarrow x = t^2 ⇒dx=2tdt\Rightarrow dx = 2t dt
∴I1=2∫sin⁡−1t⋅tdt\therefore I_1 = 2 \int \sin^{-1} t \cdot t dt
=2[sin⁡−1t⋅∫tdt−∫(ddt(sin⁡−1t)⋅∫tdt)dt]+C= 2 \left[ \sin^{-1} t \cdot \int t dt - \int \left( \frac{d}{dt} (\sin^{-1} t) \cdot \int t dt \right) dt \right] + C
=2[sin⁡−1t⋅t22−∫11−t2⋅t22dt]+C= 2 \left[ \sin^{-1} t \cdot \frac{t^2}{2} - \int \frac{1}{\sqrt{1-t^2}} \cdot \frac{t^2}{2} dt \right] + C
=t2sin⁡−1t−∫1−t2−11−t2⋅t22dt+C= t^2 \sin^{-1} t - \int \frac{1-t^2-1}{\sqrt{1-t^2}} \cdot \frac{t^2}{2} dt + C
=t2sin⁡−1t+∫1−t2dt−∫11−t2dt+C= t^2 \sin^{-1} t + \int \sqrt{1-t^2} dt - \int \frac{1}{\sqrt{1-t^2}} dt + C
=t2sin⁡−1t+(t21−t2+12sin⁡−1t)−sin⁡−1t+C= t^2 \sin^{-1} t + \left( \frac{t}{2} \sqrt{1-t^2} + \frac{1}{2} \sin^{-1} t \right) - \sin^{-1} t + C
=xsin⁡−1x+x1−x2−12sin⁡−1x+C= x \sin^{-1} \sqrt{x} + \frac{\sqrt{x} \sqrt{1-x}}{2} - \frac{1}{2} \sin^{-1} \sqrt{x} + C
=(x−12)sin⁡−1x+x1−x2+C= \left(x - \frac{1}{2}\right) \sin^{-1} \sqrt{x} + \frac{\sqrt{x} \sqrt{1-x}}{2} + C
∴ From (i), we have
I=4π(2x−12)sin⁡−1x+2x1−xπ−x+CI = \frac{4}{\pi} \left( \frac{2x-1}{2} \right) \sin^{-1} \sqrt{x} + \frac{2\sqrt{x} \sqrt{1-x}}{\pi} - x + C
=2(2x−1)πsin⁡−1x+2x1−xπ−x+C= \frac{2(2x-1)}{\pi} \sin^{-1} \sqrt{x} + \frac{2\sqrt{x} \sqrt{1-x}}{\pi} - x + C
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