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प्रश्न
उत्तर
\[\text{We have}, \]
\[I = \int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\text{ Putting x }= \cos \theta\]
\[ \Rightarrow dx = - \sin \text{ θ dθ}\]
\[ \therefore I = \int \tan^{- 1} \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \left( - \sin \text{ θ dθ} \right)\]
\[ = \int \tan^{- 1} \sqrt{\frac{2 \sin^2 \frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}}} \left( - \sin \theta \right)d\theta\]
\[ = \int \tan^{- 1} \left( \tan \frac{\theta}{2} \right) \left( - \sin \theta \right)d\theta\]
\[ = - \frac{1}{2}\int\theta \sin \theta d\theta\]
\[\text{Considering} \text{ θ as first function and} \sin \text{ θ as second function}\]
\[I = - \frac{1}{2}\left[ \theta\left( - \cos \theta \right) - \int1\left( - \cos \theta \right)d\theta \right]\]
\[ = - \frac{1}{2}\left( \theta\left( - \cos \theta \right) + \int\cos \theta d\theta \right)\]
\[ = - \frac{1}{2}\left( - \theta \cos \theta + \sin \theta \right) + C\]
\[ = - \frac{1}{2}\left[ - \theta \cos \theta + \sqrt{1 - \cos^2 \theta} \right] + C\]
\[ = - \frac{1}{2}\left[ - \cos^{- 1} x \times x + \sqrt{1 - x^2} \right] + C\]
\[ = \frac{1}{2}\left[ x \cos^{- 1} x - \sqrt{1 - x^2} \right] + C\]
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संबंधित प्रश्न
\[\int\frac{x^2 + 5x + 2}{x + 2} dx\]
If \[\int\frac{1}{5 + 4 \sin x} dx = A \tan^{- 1} \left( B \tan\frac{x}{2} + \frac{4}{3} \right) + C,\] then
\[\int\sqrt{\frac{x}{1 - x}} dx\] is equal to
\[\int\frac{1}{\sqrt{x} + \sqrt{x + 1}} \text{ dx }\]
