Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\text{Let I } = \int\frac{dx}{\sqrt{\left( x - \alpha \right) \left( \beta - x \right)}}\]
\[ = \int\frac{dx}{\sqrt{\ β x - x^2 - αβ+ α x }}\]
\[ = \int\frac{dx}{\sqrt{- x^2 + \left( \alpha + \beta \right) x - \alpha\beta}}\]
\[ = \int\frac{dx}{\sqrt{- \left[ x^2 - \left( \alpha + \beta \right) x + \alpha\beta \right]}}\]
\[ = \int\frac{dx}{\sqrt{- \left[ x^2 - \left( \alpha + \beta \right) x + \left( \frac{\alpha + \beta}{2} \right)^2 - \left( \frac{\alpha + \beta}{2} \right)^2 + \alpha\beta \right]}}\]
\[ = \int\frac{dx}{\sqrt{- \left\{ x - \left( \frac{\alpha + \beta}{2} \right) \right\}^2 + \left( \frac{\alpha + \beta}{2} \right)^2 - \alpha\beta}}\]
\[ = \int\frac{dx}{\sqrt{- \left[ x - \left( \frac{\alpha + \beta}{2} \right) \right]^2 + \frac{\left( \alpha + \beta \right)^2 - 4\alpha\beta}{4}}}\]
\[ = \int\frac{dx}{\sqrt{- \left[ x - \left( \frac{\alpha + \beta}{2} \right) \right]^2 + \left( \frac{\alpha - \beta}{2} \right)^2}}\]
\[ = \int\frac{dx}{\sqrt{\left( \frac{\alpha - \beta}{2} \right)^2 - \left( x - \left( \frac{\alpha + \beta}{2} \right) \right)^2}}\]
\[ = \sin^{- 1} \left[ \frac{x - \left( \frac{\alpha + \beta}{2} \right)}{\frac{\alpha - \beta}{2}} \right] + C\]
\[ = \sin^{- 1} \left( \frac{2x - \alpha - \beta}{\alpha - \beta} \right) + C\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integrals:
The value of \[\int\frac{\sin x + \cos x}{\sqrt{1 - \sin 2x}} dx\] is equal to
\[\int\sin x \sin 2x \text{ sin 3x dx }\]
