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प्रश्न
\[\int\frac{1}{\sqrt{\left( 1 - x^2 \right)\left\{ 9 + \left( \sin^{- 1} x \right)^2 \right\}}} dx\]
योग
उत्तर
\[\int\frac{dx}{\sqrt{\left( 1 - x^2 \right) \left( 9 + \left( \sin^{- 1} x \right)^2 \right)}}\]
\[\text{ let } \sin^{- 1} x = t\]
\[ \Rightarrow \frac{1}{\sqrt{1 - x^2}} dx = dt\]
\[Now, \int\frac{dx}{\sqrt{\left( 1 - x^2 \right) \left( 9 + \left( \sin^{- 1} x \right)^2 \right)}} \]
\[ = \int\frac{dt}{\sqrt{9 + t^2}}\]
\[ = \int\frac{dt}{\sqrt{3^2 + t^2}}\]
\[ = \text{ log } \left| t + \sqrt{3^2 + t^2} \right| + C\]
\[ = \text{ log }\left| \sin^{- 1} x + \sqrt{9 + \left( \sin^{- 1} x \right)^2} \right| + C\]
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