Advertisements
Advertisements
प्रश्न
\[\int\frac{1}{\sqrt{2x - x^2}} dx\]
योग
Advertisements
उत्तर
\[\int\frac{dx}{\sqrt{2x - x^2}}\]
\[ = \int\frac{dx}{\sqrt{2x - x^2 - 1 + 1}}\]
\[ = \int\frac{dx}{\sqrt{1 - \left( x^2 - 2x + 1 \right)}}\]
\[ = \int\frac{dx}{\sqrt{1 - \left( x - 1 \right)^2}} \]
\[ = \sin^{- 1} \left( x - 1 \right) + C \left[ \because \int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{- 1} \left( \frac{x}{a} \right) + C \right]\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\left( \frac{m}{x} + \frac{x}{m} + m^x + x^m + mx \right) dx\]
\[\int\left( 5x + 3 \right) \sqrt{2x - 1} dx\]
\[\int\frac{- \sin x + 2 \cos x}{2 \sin x + \cos x} dx\]
\[\int\frac{\cos 4x - \cos 2x}{\sin 4x - \sin 2x} dx\]
` ∫ tan 2x tan 3x tan 5x dx `
\[\int\left( 4x + 2 \right)\sqrt{x^2 + x + 1} \text{dx}\]
\[\int x^3 \cos x^4 dx\]
\[\int\frac{e^{2x}}{1 + e^x} dx\]
\[\int\frac{x^2}{\sqrt{1 - x}} dx\]
\[\int\frac{x^4 + 1}{x^2 + 1} dx\]
\[\int\frac{\sin 2x}{\sqrt{\cos^4 x - \sin^2 x + 2}} dx\]
\[\int\frac{\left( x - 1 \right)^2}{x^2 + 2x + 2} dx\]
\[\int\frac{1}{4 + 3 \tan x} dx\]
\[\int x^2 e^{- x} \text{ dx }\]
\[\int\frac{\log x}{x^n}\text{ dx }\]
\[\int x \sin x \cos x\ dx\]
\[\int x\left( \frac{\sec 2x - 1}{\sec 2x + 1} \right) dx\]
\[\int\left( e^\text{log x} + \sin x \right) \text{ cos x dx }\]
\[\int\left( \tan^{- 1} x^2 \right) x\ dx\]
\[\int\frac{x^2 \sin^{- 1} x}{\left( 1 - x^2 \right)^{3/2}} \text{ dx }\]
\[\int\left( x - 2 \right) \sqrt{2 x^2 - 6x + 5} \text{ dx }\]
\[\int\frac{x^2 + 1}{x\left( x^2 - 1 \right)} dx\]
\[\int\frac{x^2 + 1}{\left( 2x + 1 \right) \left( x^2 - 1 \right)} dx\]
Write the anti-derivative of \[\left( 3\sqrt{x} + \frac{1}{\sqrt{x}} \right) .\]
\[\int\frac{e^x \left( 1 + x \right)}{\cos^2 \left( x e^x \right)} dx =\]
\[\int\frac{1}{\sqrt{x} + \sqrt{x + 1}} \text{ dx }\]
\[\int\frac{x^3}{\left( 1 + x^2 \right)^2} \text{ dx }\]
\[\int \sin^5 x\ dx\]
\[\int\frac{1}{4 x^2 + 4x + 5} dx\]
\[\int\frac{1}{3 x^2 + 13x - 10} \text{ dx }\]
\[\int\frac{1}{5 - 4 \sin x} \text{ dx }\]
\[\int {cosec}^4 2x\ dx\]
\[\int \sin^{- 1} \sqrt{x}\ dx\]
\[\int\frac{x^2 - 2}{x^5 - x} \text{ dx}\]
\[\int\frac{\sin 4x - 2}{1 - \cos 4x} e^{2x} \text{ dx}\]
Evaluate : \[\int\frac{\cos 2x + 2 \sin^2 x}{\cos^2 x}dx\] .
\[\int\frac{x^2 + 1}{x^2 - 5x + 6} \text{ dx }\]
\[\int\frac{x^2}{x^2 + 7x + 10}\text{ dx }\]
