Advertisements
Advertisements
प्रश्न
\[\int\frac{\cos x}{1 - \cos x} \text{dx }or \int\frac{\cot x}{\text{cosec } {x }- \cot x} dx\]
बेरीज
Advertisements
उत्तर
\[\int\frac{\cot x}{\text{cosec x }- \cot x}dx\]
\[ = \int\frac{\frac{\cos x}{\sin x}}{\frac{1}{\sin x} - \frac{\cos x}{\sin x}}dx\]
\[ = \int\left( \frac{\cos x}{1 - \cos x} \right) \times \frac{\left( 1 + \cos x \right)}{\left( 1 + \cos x \right)}dx\]
\[ = \int\left( \frac{\cos x + \cos^2 x}{1 - \cos^2 x} \right)dx\]
\[ = \int\left( \frac{\cos x + \cos^2 x}{\sin^2 x} \right) dx\]
\[ = \int\left( \frac{\cos x}{\sin x} \times \frac{1}{\sin x} + \frac{\cos^2 x}{\sin^2 x} \right)dx\]
\[ = \int\left[ \left( \text{cot x cosec x} \right) + \cot^2 x \right]dx\]
\[ = \int\left[ \text{cosec x cot x }+ {cosec}^2 x - 1 \right]dx\]
\[ = -\text{ cosec x} - \cot x - x + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int \left( a \tan x + b \cot x \right)^2 dx\]
Write the primitive or anti-derivative of
\[f\left( x \right) = \sqrt{x} + \frac{1}{\sqrt{x}} .\]
\[\int \left( 2x - 3 \right)^5 + \sqrt{3x + 2} \text{dx} \]
\[\int\left( 5x + 3 \right) \sqrt{2x - 1} dx\]
\[\int\frac{1 - \cot x}{1 + \cot x} dx\]
\[\int\frac{1}{\sqrt{1 - x^2}\left( 2 + 3 \sin^{- 1} x \right)} dx\]
\[\int\frac{\tan x}{\sqrt{\cos x}} dx\]
\[\int\frac{\cos x - \sin x}{1 + \sin 2x} dx\]
\[\int\frac{x \sin^{- 1} x^2}{\sqrt{1 - x^4}} dx\]
\[\int \cot^6 x \text{ dx }\]
\[\int \sin^5 x \cos x \text{ dx }\]
\[\int \sin^3 x \cos^6 x \text{ dx }\]
\[\int \sin^3 x \cos^5 x \text{ dx }\]
\[\int\frac{3 x^5}{1 + x^{12}} dx\]
\[\int\frac{1}{\sqrt{7 - 3x - 2 x^2}} dx\]
\[\int\frac{\cos x - \sin x}{\sqrt{8 - \sin2x}}dx\]
\[\int\frac{x^2 \left( x^4 + 4 \right)}{x^2 + 4} \text{ dx }\]
\[\int\frac{x + 1}{\sqrt{x^2 + 1}} dx\]
\[\int\frac{2}{2 + \sin 2x}\text{ dx }\]
\[\int\frac{1}{3 + 2 \cos^2 x} \text{ dx }\]
\[\int\frac{1}{1 - \tan x} \text{ dx }\]
\[\int\frac{1}{3 + 4 \cot x} dx\]
\[\int\text{ log }\left( x + 1 \right) \text{ dx }\]
\[\int x \cos^2 x\ dx\]
\[\int x \sin x \cos x\ dx\]
\[\int\frac{x^3 \sin^{- 1} x^2}{\sqrt{1 - x^4}} \text{ dx }\]
\[\int e^x \left( \frac{1}{x^2} - \frac{2}{x^3} \right) dx\]
\[\int\left( \frac{1}{\log x} - \frac{1}{\left( \log x \right)^2} \right) dx\]
\[\int\sqrt{x^2 - 2x} \text{ dx}\]
\[\int\left( x + 1 \right) \sqrt{x^2 + x + 1} \text{ dx }\]
\[\int\frac{x^2}{\left( x^2 + 1 \right) \left( 3 x^2 + 4 \right)} dx\]
\[\int\frac{1}{\sin x + \sin 2x} dx\]
\[\int\frac{1}{\left( 2 x^2 + 3 \right) \sqrt{x^2 - 4}} \text{ dx }\]
\[\int\frac{x}{4 + x^4} \text{ dx }\] is equal to
\[\int\sqrt{\sin x} \cos^3 x\ \text{ dx }\]
\[\int\frac{\sin^6 x}{\cos x} \text{ dx }\]
\[\int\frac{\log x}{x^3} \text{ dx }\]
\[\int\frac{1}{x \sqrt{1 + x^n}} \text{ dx}\]
\[\int\frac{x \sin^{- 1} x}{\left( 1 - x^2 \right)^{3/2}} \text{ dx}\]
\[\int\frac{e^{m \tan^{- 1} x}}{\left( 1 + x^2 \right)^{3/2}} \text{ dx}\]
