Advertisements
Advertisements
प्रश्न
\[\int\frac{1}{\left( \sin^{- 1} x \right) \sqrt{1 - x^2}} \text{ dx} \]
योग
उत्तर
\[\text{ Let I} = \int\frac{1}{\sin^{- 1} x \cdot \sqrt{1 - x^2}}dx\]
\[\text{ Putting sin}^{- 1} x = t\]
\[ \Rightarrow \frac{dx}{\sqrt{1 - x^2}} = dt\]
\[ \therefore I = \int\frac{dt}{t}\]
\[ = \text{ ln }\left| t \right| + C\]
` = \text{ ln } | sin ^-1 x| + c ( ∵ t = sin ^-1 x ) `
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\sqrt{x}\left( 3 - 5x \right) dx\]
\[\int\frac{\tan x}{\sec x + \tan x} dx\]
\[\int\frac{\cos x}{1 + \cos x} dx\]
\[\int\frac{1}{1 - \sin\frac{x}{2}} dx\]
\[\int \tan^2 \left( 2x - 3 \right) dx\]
Integrate the following integrals:
\[\int\text{sin 2x sin 4x sin 6x dx} \]
\[\int\frac{\sin 2x}{\sin \left( x - \frac{\pi}{6} \right) \sin \left( x + \frac{\pi}{6} \right)} dx\]
\[\int\left( \frac{x + 1}{x} \right) \left( x + \log x \right)^2 dx\]
\[\int 5^{5^{5^x}} 5^{5^x} 5^x dx\]
\[\int\frac{1}{\sin^3 x \cos^5 x} dx\]
\[\int\frac{1}{\sqrt{\left( x - \alpha \right)\left( \beta - x \right)}} dx, \left( \beta > \alpha \right)\]
\[\int\frac{x + 7}{3 x^2 + 25x + 28}\text{ dx}\]
\[\int\frac{x^2 + 1}{x^2 - 5x + 6} dx\]
\[\int\frac{x + 2}{\sqrt{x^2 - 1}} \text{ dx }\]
\[\int\frac{1}{\cos x \left( \sin x + 2 \cos x \right)} dx\]
\[\int\frac{1}{p + q \tan x} \text{ dx }\]
\[\int x^2 \sin^2 x\ dx\]
\[\int \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int x^2 \sin^{- 1} x\ dx\]
\[\int \cos^{- 1} \left( 4 x^3 - 3x \right) \text{ dx }\]
\[\int \cos^3 \sqrt{x}\ dx\]
\[\int x^2 \sqrt{a^6 - x^6} \text{ dx}\]
\[\int\sqrt{x^2 - 2x} \text{ dx}\]
\[\int\frac{x^2}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)} dx\]
\[\int\frac{5 x^2 - 1}{x \left( x - 1 \right) \left( x + 1 \right)} dx\]
\[\int\frac{1}{\sin x + \sin 2x} dx\]
\[\int\frac{1}{\cos x + \sqrt{3} \sin x} \text{ dx } \] is equal to
\[\int\frac{\sin^2 x}{\cos^4 x} dx =\]
\[\int\frac{\left( 2^x + 3^x \right)^2}{6^x} \text{ dx }\]
\[\int\frac{\sin x}{1 + \sin x} \text{ dx }\]
\[\int\frac{x^2}{\left( x - 1 \right)^3} dx\]
\[\int\sqrt{\frac{1 + x}{x}} \text{ dx }\]
\[\int\frac{1}{a + b \tan x} \text{ dx }\]
\[\int\frac{1}{\sin^4 x + \cos^4 x} \text{ dx}\]
\[\int\frac{\sin^2 x}{\cos^6 x} \text{ dx }\]
\[\int\sqrt{1 + 2x - 3 x^2}\text{ dx } \]
\[\int\frac{x^2}{x^2 + 7x + 10} dx\]
Find: `int (3x +5)/(x^2+3x-18)dx.`
