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NCERT Class XII Mathematics Chapter - - Solutions

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Exercise - 7.6
Integrate the functions in Exercises 134 to 155.
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Question : 155 of 271
Marks: +1, -0
sin⁡−1(2x1+x2)\sin^{-1}\left(\frac{2x}{1+x^2}\right)
Solution:  
Let I=∫sin⁡−1(2x1+x2)dxI=\int \sin^{-1}\left(\frac{2x}{1+x^2}\right)dx
Putx x=tan⁡t⇒dx=sec⁡2t dtx=\tan t \Rightarrow dx = \sec^2 t\,dt
∴I=∫sin⁡−1(2tan⁡t1+tan⁡2t)sec⁡2t dt\therefore I=\int \sin^{-1}\left(\frac{2\tan t}{1+\tan^2 t}\right)\sec^2 t\,dt
=∫sin⁡−1(sin⁡2t)sec⁡2t dt=\int \sin^{-1}(\sin 2t)\sec^2 t\,dt
=∫2tsec⁡2t dt=\int 2t\sec^2 t\,dt
=2[t∫sec⁡2t dt−∫(ddtt⋅∫sec⁡2t dt)dt]=2\left[t\int \sec^2 t\,dt - \int\left(\frac{d}{dt}t \cdot \int \sec^2 t\,dt\right)dt\right]
=2[ttan⁡t−∫1⋅tan⁡t dt]=2\left[t\tan t - \int 1\cdot\tan t\,dt\right]
=2ttan⁡t+2log⁡∣cos⁡t∣+C=2t\tan t + 2\log\left|\cos t\right| + C
=2tan⁡−1x⋅x+2log⁡∣11=x2∣+C=2\tan^{-1}x \cdot x + 2\log\left|\frac{1}{\sqrt{1=x^2}}\right| + C
=2xtan⁡−1x+2log⁡∣(1+x2)−1/2∣+C=2x\tan^{-1}x + 2\log\left|(1+x^2)^{-1/2}\right| + C
=2xtan⁡−1x+2(−12)log⁡∣1+x2∣+C=2x\tan^{-1}x + 2\left(-\frac{1}{2}\right)\log\left|1+x^2\right| + C
=2xtan⁡−1x−log⁡∣1+x2∣+C=2x\tan^{-1}x - \log\left|1+x^2\right| + C
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