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NCERT Class XII Mathematics Chapter - - Solutions

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Prove the following (Exercises 262 to 267)
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Question : 267 of 271
Marks: +1, -0
∫01sin⁡−1x dx=π2−1\int\limits_{0}^{1}\sin^{-1}x\,dx=\frac{\pi}{2}-1
Solution:  
∫01sin⁡−1x dx\int\limits_{0}^{1}\sin^{-1}x\,dx
=∫01sin⁡−1x⋅1 dx=\int\limits_{0}^{1}\sin^{-1}x\cdot 1\,dx
Integrating by parts
[sin⁡−1x∫dx−∫(ddx(sin⁡−1x)⋅∫1 dx)dx]01\left[ \sin^{-1}x\int dx-\int\left(\frac{d}{dx}(\sin^{-1}x)\cdot\int 1\,dx\right)dx \right]_{0}^{1}
[sin⁡−1x⋅x−∫11−x2×x dx]01\left[ \sin^{-1}x\cdot x-\int\frac{1}{\sqrt{1-x^{2}}}\times x\,dx \right]_{0}^{1}
[sin⁡−1x⋅x+12∫−2x1−x2 dx]01\left[ \sin^{-1}x\cdot x+\frac{1}{2}\int\frac{-2x}{\sqrt{1-x^{2}}}\,dx \right]_{0}^{1}
[xsin⁡−1x+12⋅(1−x2)1/21/2]01\left[ x\sin^{-1}x+\frac{1}{2}\cdot\frac{(1-x^{2})^{1/2}}{1/2} \right]_{0}^{1}
[(sin⁡−11)+12×21(0−1)]\left[ (\sin^{-1}1)+\frac{1}{2}\times\frac{2}{1}(0-1) \right]
=π2−1=\frac{\pi}{2}-1
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