मराठी

∫ 1 √ 1 − X 2 ( Sin − 1 X ) 2 D X

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प्रश्न

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

बेरीज
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उत्तर

 

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

 

\[Let, \sin^{- 1} x = t\]

\[ \Rightarrow \frac{1}{\sqrt{1 - x^2}} = \frac{dt}{dx}\]

\[ \Rightarrow \frac{1}{\sqrt{1 - x^2}} dx = dt\]

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

\[ = \int\frac{dt}{t^2}\]

\[ = \int t^{- 2} dt\]

\[ = \frac{t^{- 2 + 1}}{- 2 + 1} + C\]

\[ = \frac{- 1}{t} + C\]

\[ = - \frac{1}{\sin^{- 1} x} + C\]

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पाठ 18: Indefinite Integrals - Exercise 19.09 [पृष्ठ ५७]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 18 Indefinite Integrals
Exercise 19.09 | Q 10 | पृष्ठ ५७
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