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NCERT Class XII Mathematics Chapter - - Solutions

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Question : 251 of 271
Marks: +1, -0
tan⁡−11−x1+x\tan^{-1}\sqrt{\frac{1-x}{1+x}}
Solution:  
Let I=∫tan⁡−11−x1+x dxI=\int \tan^{-1}\sqrt{\frac{1-x}{1+x}} \, dx
Let x=cos⁡θx = \cos \theta ⇒dx=−sin⁡θ dθ\Rightarrow dx = -\sin\theta \, d\theta
∴I=∫tan⁡−11−cos⁡θ1+cos⁡θ(−sin⁡θ) dθ\therefore I = \int \tan^{-1}\sqrt{\frac{1-\cos\theta}{1+\cos\theta}} (-\sin\theta) \, d\theta
=−∫tan⁡−1(tan⁡θ2)(sin⁡θ) dθ=-\int \tan^{-1}\left(\frac{\tan\theta}{2}\right) (\sin\theta) \, d\theta
=−∫(θ2)(sin⁡θ) dθ=-\int \left(\frac{\theta}{2}\right)(\sin\theta) \, d\theta
=−12[θ∫sin⁡θ dθ−∫(ddθ(θ)∫sin⁡θ dθ)dθ]=-\frac{1}{2} \left[ \theta \int \sin\theta \, d\theta - \int \left( \frac{d}{d\theta}(\theta) \int \sin\theta \, d\theta \right) d\theta \right]
=−12[θ(−cos⁡θ)−∫1⋅(−cos⁡θ) dθ]=-\frac{1}{2} \left[ \theta (-\cos\theta) - \int 1 \cdot (-\cos\theta) \, d\theta \right]
=12θcos⁡θ−12∫cos⁡θ dθ=\frac{1}{2}\theta \cos\theta - \frac{1}{2} \int \cos\theta \, d\theta
=12θcos⁡θ−12sin⁡θ+C=\frac{1}{2}\theta \cos\theta - \frac{1}{2} \sin\theta + C
=12θcos⁡θ−121−cos⁡2θ+C=\frac{1}{2}\theta \cos\theta - \frac{1}{2} \sqrt{1 - \cos^2\theta} + C
=12[xcos⁡−1x−1−x2]+C=\frac{1}{2} \left[ x \cos^{-1} x - \sqrt{1 - x^2} \right] + C
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