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∫ { E Sin − 1 X } 2 √ 1 − X 2 D X - Mathematics

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प्रश्न

\[\int\frac{\left\{ e^{\sin^{- 1} }x \right\}^2}{\sqrt{1 - x^2}} dx\]

योग
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उत्तर

\[\int\frac{\left( e^\{sin^{- 1} x \right)^2}{\sqrt{1 - x^2}} dx\]

 

` Let     e^{sin-1 _x }= t `

 Differentiating both sides   w . r . t . x, 

`e^{sin-1 _x } ×  1 / \sqrt{ 1 - x^2 }    ` dx  = dt 

  `  Now  , ∫  (e^{sin-1  _ x }) ^2/ \sqrt{1-x^2}   ` dx

  `  ∫  e^{sin-1    _x }  .   {e^{sin-1  _ x }}/ \sqrt{1-x^2}   ` dx

`  ∫  t . dt 

\[ = \frac{t^2}{2} + C\]

`  (e^{sin-1    _x })^2 /2 + C`

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अध्याय 19: Indefinite Integrals - Exercise 19.09 [पृष्ठ ५७]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.09 | Q 8 | पृष्ठ ५७

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