Advertisements
Advertisements
प्रश्न
\[\int\frac{1}{\sqrt{1 - x^2} \left( \sin^{- 1} x \right)^2} dx\]
उत्तर
\[\int\frac{dx}{\sqrt{1 - x^2} \left( \sin^{- 1} x \right)^2}\]
\[Let, \sin^{- 1} x = t\]
\[ \Rightarrow \frac{1}{\sqrt{1 - x^2}} = \frac{dt}{dx}\]
\[ \Rightarrow \frac{1}{\sqrt{1 - x^2}} dx = dt\]
\[Now, \int\frac{dx}{\sqrt{1 - x^2} \left( \sin^{- 1} x \right)^2}\]
\[ = \int\frac{dt}{t^2}\]
\[ = \int t^{- 2} dt\]
\[ = \frac{t^{- 2 + 1}}{- 2 + 1} + C\]
\[ = \frac{- 1}{t} + C\]
\[ = - \frac{1}{\sin^{- 1} x} + C\]
APPEARS IN
संबंधित प्रश्न
Evaluate : `int_0^3dx/(9+x^2)`
` ∫ cot^3 x "cosec"^2 x dx `
\[\int\frac{\left\{ e^{\sin^{- 1} }x \right\}^2}{\sqrt{1 - x^2}} dx\]
\[\int\frac{\cot x}{\sqrt{\sin x}} dx\]
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate:
Evaluate:\[\int\frac{\sin \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate: \[\int 2^x \text{ dx }\]
Evaluate: \[\int\left( 1 - x \right)\sqrt{x}\text{ dx }\]
Evaluate: \[\int\frac{x + \cos6x}{3 x^2 + \sin6x}\text{ dx }\]
Evaluate: \[\int\frac{2}{1 - \cos2x}\text{ dx }\]
Evaluate:
Evaluate the following:
`int sqrt(5 - 2x + x^2) "d"x`
Evaluate the following:
`int sqrt(x)/(sqrt("a"^3 - x^3)) "d"x`
