Advertisements
Advertisements
प्रश्न
\[\int\frac{1}{1 - \cos x} dx\]
बेरीज
उत्तर
\[\int\frac{dx}{1 - \cos x}\]
\[ = \int\frac{dx}{1 - \cos x} \times \frac{1 + \ cosx}{1 + \ cosx}\]
\[ = \int\left( \frac{1 + \cos x}{1 - \cos^2 x} \right)dx\]
\[ = \int\left( \frac{1 + \cos x}{\sin^2 x} \right)dx\]
\[ = \int\left( \frac{1}{\sin^2 x} + \frac{\cos x}{\sin x} \times \frac{1}{\sin x} \right)dx\]
\[ = \int\left( {cosec}^2 x + \text{cosec x }\cot x \right)dx\]
\[ = - \cot x - \text{cosec x} + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{\left( x + 1 \right)\left( x - 2 \right)}{\sqrt{x}} dx\]
\[\int\frac{\tan x}{\sec x + \tan x} dx\]
\[\int\frac{1}{1 + \cos 2x} dx\]
\[\int\frac{2x - 1}{\left( x - 1 \right)^2} dx\]
\[\int\frac{3x + 5}{\sqrt{7x + 9}} dx\]
` ∫ cos 3x cos 4x` dx
\[\int\sqrt{\frac{1 - \sin 2x}{1 + \sin 2x}} dx\]
\[\int\frac{\sec^2 x}{\tan x + 2} dx\]
\[\int\frac{\sin 2x}{\sin 5x \sin 3x} dx\]
\[\int\frac{\cos x - \sin x}{1 + \sin 2x} dx\]
` ∫ x {tan^{- 1} x^2}/{1 + x^4} dx`
\[\int \cot^5 x \text{ dx }\]
` ∫ {1}/{a^2 x^2- b^2}dx`
\[\int\frac{1}{\sqrt{1 + 4 x^2}} dx\]
\[\int\frac{1}{\sqrt{\left( 2 - x \right)^2 - 1}} dx\]
\[\int\frac{x}{x^4 + 2 x^2 + 3} dx\]
\[\int\frac{1}{\sqrt{5 x^2 - 2x}} dx\]
\[\int\frac{\cos x}{\sqrt{4 + \sin^2 x}} dx\]
\[\int\frac{x}{\sqrt{4 - x^4}} dx\]
\[\int\frac{6x - 5}{\sqrt{3 x^2 - 5x + 1}} \text{ dx }\]
\[\int\frac{1}{\sqrt{3} \sin x + \cos x} dx\]
\[\int\frac{\log x}{x^n}\text{ dx }\]
` ∫ sin x log (\text{ cos x ) } dx `
\[\int\frac{x^2 \sin^{- 1} x}{\left( 1 - x^2 \right)^{3/2}} \text{ dx }\]
\[\int\left( \frac{1}{\log x} - \frac{1}{\left( \log x \right)^2} \right) dx\]
\[\int\frac{x^2 + x - 1}{x^2 + x - 6} dx\]
\[\int\frac{5 x^2 + 20x + 6}{x^3 + 2 x^2 + x} dx\]
\[\int\frac{1}{\left( x + 1 \right)^2 \left( x^2 + 1 \right)} dx\]
\[\int\frac{x}{4 + x^4} \text{ dx }\] is equal to
\[\int\frac{x^9}{\left( 4 x^2 + 1 \right)^6}dx\] is equal to
\[\int\frac{1 - x^4}{1 - x} \text{ dx }\]
\[\int\frac{1}{e^x + 1} \text{ dx }\]
\[\int\frac{1}{\text{ sin} \left( x - a \right) \text{ sin } \left( x - b \right)} \text{ dx }\]
\[\int\frac{1}{5 - 4 \sin x} \text{ dx }\]
\[\int x\sqrt{1 + x - x^2}\text{ dx }\]
\[\int\frac{\log \left( 1 - x \right)}{x^2} \text{ dx}\]
\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2} \text{ dx}\]
Find : \[\int\frac{dx}{\sqrt{3 - 2x - x^2}}\] .
\[\int\frac{x^2}{x^2 + 7x + 10}\text{ dx }\]