Advertisements
Advertisements
प्रश्न
\[\int\frac{1}{1 - \sin x} dx\]
योग
उत्तर
\[\int\frac{dx}{1 - \sin x}\]
\[ = \int\frac{\left( 1 + \sin x \right)}{\left( 1 - \sin x \right) \times \left( 1 + \sin x \right)}dx\]
\[ = \int\left( \frac{1 + \sin x}{1 - \sin^2 x} \right)dx\]
\[ = \int\left( \frac{1 + \sin x}{\cos^2 x} \right)dx\]
\[ = \int\left( \frac{1}{\cos^2 x} + \frac{\sin x}{\cos x} \times \frac{1}{\cos x} \right)dx\]
\[ = \int\left( \sec^2 x + \sec x \tan x \right)dx\]
\[ = \tan x + \sec x + C\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{x^5 + x^{- 2} + 2}{x^2} dx\]
\[\int\frac{1}{1 + \cos 2x} dx\]
\[\int \cot^{- 1} \left( \frac{\sin 2x}{1 - \cos 2x} \right) dx\]
\[\int\frac{x^2 + 5x + 2}{x + 2} dx\]
\[\int\frac{\text{sin} \left( x - \alpha \right)}{\text{sin }\left( x + \alpha \right)} dx\]
\[\int\frac{1}{x (3 + \log x)} dx\]
\[\int\frac{1}{\sqrt{1 - x^2}\left( 2 + 3 \sin^{- 1} x \right)} dx\]
\[\int\frac{2 \cos 2x + \sec^2 x}{\sin 2x + \tan x - 5} dx\]
\[\int\frac{\sin 2x}{\sin \left( x - \frac{\pi}{6} \right) \sin \left( x + \frac{\pi}{6} \right)} dx\]
\[\int\frac{\sin \left( \tan^{- 1} x \right)}{1 + x^2} dx\]
\[\int\frac{e^{m \tan^{- 1} x}}{1 + x^2} dx\]
\[\int\frac{1}{\sqrt{x} + x} \text{ dx }\]
\[\int \sec^4 2x \text{ dx }\]
\[\int\frac{1}{\sqrt{7 - 3x - 2 x^2}} dx\]
\[\int\frac{1}{\sqrt{16 - 6x - x^2}} dx\]
\[\int\frac{\sin 2x}{\sqrt{\cos^4 x - \sin^2 x + 2}} dx\]
\[\int\frac{x + 2}{\sqrt{x^2 + 2x - 1}} \text{ dx }\]
\[\int\frac{\cos x}{\cos 3x} \text{ dx }\]
\[\int\frac{1}{5 + 7 \cos x + \sin x} dx\]
\[\int x \cos x\ dx\]
\[\int\frac{\sin^{- 1} x}{x^2} \text{ dx }\]
\[\int \sin^{- 1} \sqrt{\frac{x}{a + x}} \text{ dx }\]
\[\int e^x \left( \frac{\sin 4x - 4}{1 - \cos 4x} \right) dx\]
\[\int e^x \left( \log x + \frac{1}{x} \right) dx\]
\[\int x^2 \sqrt{a^6 - x^6} \text{ dx}\]
\[\int\sqrt{\cot \text{θ} d } \text{ θ}\]
\[\int\frac{1}{\left( x - 1 \right) \sqrt{x + 2}} \text{ dx }\]
\[\int\frac{1}{\left( x^2 + 1 \right) \sqrt{x}} \text{ dx }\]
\[\int\frac{x}{\left( x^2 + 2x + 2 \right) \sqrt{x + 1}} \text{ dx}\]
Write the anti-derivative of \[\left( 3\sqrt{x} + \frac{1}{\sqrt{x}} \right) .\]
\[\int\frac{x^3}{x + 1}dx\] is equal to
\[\int \sec^2 x \cos^2 2x \text{ dx }\]
\[\int \text{cosec}^2 x \text{ cos}^2 \text{ 2x dx} \]
\[\int\frac{e^x - 1}{e^x + 1} \text{ dx}\]
\[\int \sin^5 x\ dx\]
\[\int\frac{1}{\left( \sin x - 2 \cos x \right) \left( 2 \sin x + \cos x \right)} \text{ dx }\]
\[\int\frac{\sin^2 x}{\cos^6 x} \text{ dx }\]
\[\int x \sec^2 2x\ dx\]
\[\int\frac{x^2 + x + 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} \text{ dx}\]
