Question

निम्नलिखित का मान निकालिए-

${\displaystyle \int {\sin }^{-1}\sqrt{\frac{x}{\text{a}+x}}\text{d}x}$  (संकेत: x = a tan²θ रखिए)

Answer 1

मान लीजिए I = ${\displaystyle \int {\sin }^{-1}\sqrt{\frac{x}{\text{a}+x}}\text{d}x}$

x = a tan²θ रखिए

dx = 2a tan θ . sec²θ . dθ

∴ I = ${\displaystyle \int {\sin }^{-1}\sqrt{\frac{\text{a}{\tan }^2\theta }{\text{a}+\text{a}{\tan }^2\theta }}\cdot 2\text{a}\tan \theta \cdot {\sec }^2\theta \text{d}\theta }$

= ${\displaystyle \int {\sin }^{-1} \frac{\sqrt{\text{a}}\tan \theta }{\sqrt{\text{a}}\tan \theta }\cdot 2\text{a}\tan \theta \cdot \sec \theta \text{d}\theta }$

= ${\displaystyle \int {\sin }^{-1} \left(\frac{\frac{\sin \theta }{\cos \theta }}{\frac{1}{\cos \theta }}\right)\cdot 2\text{a}\tan \theta \cdot {\sec }^2\theta \text{d}\theta }$

= ${\displaystyle \int {\sin }^{-1}\left(\sin \theta \right)\cdot 2\text{a}\tan \theta \cdot {\sec }^2\theta \text{d}\theta }$

= ${\displaystyle 2\text{a}\int \theta \tan \theta \cdot {\sec }^2\theta \text{d}\theta }$

= ${\displaystyle 2\text{a}[\theta \int \tan \theta \cdot {\sec }^2\theta \text{d}\theta -\int [\text{D}\left(\theta \right)\cdot \int \tan \theta \cdot {\sec }^2\theta \text{d}\theta ]]}$

= ${\displaystyle 2\text{a}[\theta \cdot \frac{{\tan }^2\theta }{2}-\int \frac{1\cdot {\tan }^2\theta }{2} \text{d}\theta ]}$

= ${\displaystyle 2\text{a}[\theta \cdot \frac{{\tan }^2\theta }{2}-\frac{1}{2}\int ({\sec }^2\theta -1)\text{d}\theta ]}$

= ${\displaystyle 2\text{a}[\theta \cdot \frac{{\tan }^2\theta }{2}-\frac{1}{2}(\tan \theta -\theta )]}$

= ${\displaystyle 2\text{a}[\theta \cdot \frac{{\tan }^2\theta }{2}-\frac{1}{2}\tan \theta +\frac{1}{2}\theta ]}$

= ${\displaystyle 2\text{a}[{\tan }^{-1}\sqrt{\frac{x}{\text{a}}}\cdot \frac{x}{2\text{a}}-\frac{1}{2}\sqrt{\frac{x}{\text{a}}}+\frac{1}{2}{\tan }^{-1}\sqrt{\frac{x}{\text{a}}}]+\text{C}}$

= ${\displaystyle \text{a}[\frac{x}{\text{a}}{\tan }^{-1}\sqrt{\frac{x}{\text{a}}}-\sqrt{\frac{x}{\text{a}}}+{\tan }^{-1}\sqrt{\frac{x}{\text{a}}}]+\text{C}}$

अत:, I =  ${\displaystyle \text{a}[\frac{x}{\text{a}}{\tan }^{-1}\sqrt{\frac{x}{\text{a}}}-\sqrt{\frac{x}{\text{a}}}+{\tan }^{-1}\sqrt{\frac{x}{\text{a}}}]+\text{C}}$