Advertisements
Advertisements
प्रश्न
\[\int \cot^{- 1} \left( \frac{\sin 2x}{1 - \cos 2x} \right) dx\]
बेरीज
Advertisements
उत्तर
\[\int \cot^{- 1} \left( \frac{\sin 2x}{1 - \cos 2x} \right)dx\]
` = ∫ cot ^-1 (( 2 sin x cos x) /( 2 sin^2 x))` dx ` [∴ sin 2x = 2 sin x cos x & 1 - cos 2x = 2 sin^2 x ]`
\[ = \int \cot^{- 1} \left( \cot x \right)dx\]
` = ∫ x dx `
\[ = \frac{x^2}{2} + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\left( 2 - 3x \right) \left( 3 + 2x \right) \left( 1 - 2x \right) dx\]
\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]
\[\int \left( 2x - 3 \right)^5 + \sqrt{3x + 2} \text{dx} \]
\[\int\sin x\sqrt{1 + \cos 2x} dx\]
\[\int\frac{1}{\sqrt{x + 3} - \sqrt{x + 2}} dx\]
\[\int\frac{e^{3x}}{e^{3x} + 1} dx\]
\[\int\frac{\sec x \tan x}{3 \sec x + 5} dx\]
\[\int\frac{- \sin x + 2 \cos x}{2 \sin x + \cos x} dx\]
\[\int\frac{1 + \cot x}{x + \log \sin x} dx\]
\[\int\frac{\log\left( 1 + \frac{1}{x} \right)}{x \left( 1 + x \right)} dx\]
\[\int\frac{\cos x - \sin x}{1 + \sin 2x} dx\]
` ∫ {1}/{a^2 x^2- b^2}dx`
` ∫ { x^2 dx}/{x^6 - a^6} dx `
\[\int\frac{1}{\sqrt{7 - 3x - 2 x^2}} dx\]
\[\int\frac{x}{\sqrt{x^4 + a^4}} dx\]
\[\int\frac{1}{x\sqrt{4 - 9 \left( \log x \right)^2}} dx\]
\[\int\frac{\cos x}{\sqrt{4 - \sin^2 x}} dx\]
` ∫ {x-3} /{ x^2 + 2x - 4 } dx `
\[\int\frac{\left( 1 - x^2 \right)}{x \left( 1 - 2x \right)} \text
{dx\]
\[\int\frac{x}{\sqrt{x^2 + x + 1}} \text{ dx }\]
\[\int\frac{1}{5 - 4 \sin x} \text{ dx }\]
\[\int\frac{1}{3 + 2 \sin x + \cos x} \text{ dx }\]
\[\int\frac{4 \sin x + 5 \cos x}{5 \sin x + 4 \cos x} \text{ dx }\]
\[\int \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) \text{ dx }\]
\[\int x^3 \tan^{- 1}\text{ x dx }\]
\[\int x \cos^3 x\ dx\]
\[\int x^2 \sqrt{a^6 - x^6} \text{ dx}\]
\[\int\frac{x^2 - 1}{x^4 + 1} \text{ dx }\]
\[\int\left( x - 1 \right) e^{- x} dx\] is equal to
\[\int\frac{\sin x}{3 + 4 \cos^2 x} dx\]
\[\int\frac{\cos 2x - 1}{\cos 2x + 1} dx =\]
\[\int\frac{\cos2x - \cos2\theta}{\cos x - \cos\theta}dx\] is equal to
\[\int \text{cosec}^2 x \text{ cos}^2 \text{ 2x dx} \]
\[\int\sqrt{\text{ cosec x} - 1} \text{ dx }\]
\[\int\frac{1}{4 \sin^2 x + 4 \sin x \cos x + 5 \cos^2 x} \text{ dx }\]
\[\int\frac{1}{1 - 2 \sin x} \text{ dx }\]
\[\int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]
\[\int\frac{x \sin^{- 1} x}{\left( 1 - x^2 \right)^{3/2}} \text{ dx}\]
