Advertisements
Advertisements
प्रश्न
` ∫ {cosec x} / {"cosec x "- cot x} ` dx
बेरीज
उत्तर
` ∫ ( {"cosec "x} / {"cosec x "- cot x} )` dx
\[ = \int\frac{\text{cosec x}\left( \text{cosec x }+ \cot x \right)}{\left(\text{ cosec x} - \cot x \right) \left( \text{cosec x }+ \cot x \right)}dx\]
\[ = \int\frac{\text{cosec x} \left( \text{cosec x }+ \cot x \right)}{\left( {\text{cosec}}^2 x - \cot^2 x \right)}dx\]
\[ = \int\left( {\text{cosec}}^2 x + \text{cosec x } \cot x \right)dx\]
\[ = - \cot x - \text{cosec x }+ C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{\cos^2 x - \sin^2 x}{\sqrt{1} + \cos 4x} dx\]
\[\int\frac{1}{1 - \cos x} dx\]
\[\int\frac{\left( x^3 + 8 \right)\left( x - 1 \right)}{x^2 - 2x + 4} dx\]
\[\int\frac{2 - 3x}{\sqrt{1 + 3x}} dx\]
` ∫ cos 3x cos 4x` dx
\[\int\frac{\text{sin} \left( x - a \right)}{\text{sin}\left( x - b \right)} dx\]
\[\int\frac{1}{1 + \sqrt{x}} dx\]
\[\int2x \sec^3 \left( x^2 + 3 \right) \tan \left( x^2 + 3 \right) dx\]
` ∫ e^{m sin ^-1 x}/ \sqrt{1-x^2} ` dx
\[\int\frac{1}{x^2 \left( x^4 + 1 \right)^{3/4}} dx\]
\[\int\frac{x^2}{\sqrt{1 - x}} dx\]
\[\int\frac{1}{x^2 - 10x + 34} dx\]
\[\int\frac{\sec^2 x}{1 - \tan^2 x} dx\]
\[\int\frac{x}{x^2 + 3x + 2} dx\]
\[\int\frac{2}{2 + \sin 2x}\text{ dx }\]
\[\int\frac{1}{\left( \sin x - 2 \cos x \right)\left( 2 \sin x + \cos x \right)} \text{ dx }\]
\[\int\frac{1}{5 + 4 \cos x} dx\]
\[\int\frac{1}{5 + 7 \cos x + \sin x} dx\]
\[\int\frac{1}{4 + 3 \tan x} dx\]
\[\int x e^{2x} \text{ dx }\]
\[\int\left( x + 1 \right) \text{ e}^x \text{ log } \left( x e^x \right) dx\]
\[\int x^2 \tan^{- 1} x\text{ dx }\]
\[\int x \sin^3 x\ dx\]
\[\int\left( x + 1 \right) \sqrt{x^2 - x + 1} \text{ dx}\]
\[\int\frac{1}{\cos x \left( 5 - 4 \sin x \right)} dx\]
\[\int\frac{1}{\sin x + \sin 2x} dx\]
\[\int\frac{x^2 + 1}{x^4 + 7 x^2 + 1} 2 \text{ dx }\]
\[\int\frac{1}{\left( 1 + x^2 \right) \sqrt{1 - x^2}} \text{ dx }\]
\[\int\frac{1}{\left( 2 x^2 + 3 \right) \sqrt{x^2 - 4}} \text{ dx }\]
Write a value of
\[\int e^{3 \text{ log x}} x^4\text{ dx}\]
\[\int\frac{x^3}{\sqrt{1 + x^2}}dx = a \left( 1 + x^2 \right)^\frac{3}{2} + b\sqrt{1 + x^2} + C\], then
\[\int \tan^4 x\ dx\]
\[\int\frac{1}{\sqrt{x^2 + a^2}} \text{ dx }\]
\[\int\sqrt{\frac{1 - x}{x}} \text{ dx}\]
\[\int\frac{\sin^6 x}{\cos x} \text{ dx }\]
\[\int\log \left( x + \sqrt{x^2 + a^2} \right) \text{ dx}\]
\[\int\frac{1}{x \sqrt{1 + x^n}} \text{ dx}\]
\[\int\frac{x^2}{\left( x - 1 \right)^3 \left( x + 1 \right)} \text{ dx}\]
