Advertisements
Advertisements
प्रश्न
` ∫ {"cosec" x }/ { log tan x/2 ` dx
बेरीज
उत्तर
` Note : "Here, we are considering " log x as log_e x `
\[\text{Let I} = \int\frac{cosec x}{\log \tan\frac{x}{2}}dx\]
\[Putting\ \log \tan \frac{x}{2} = t\]
\[ \Rightarrow \frac{1}{2}\frac{\sec^2 \frac{x}{2}}{\tan\frac{x}{2}} = \frac{dt}{dx}\]
\[ \Rightarrow \frac{1}{2 \sin\frac{x}{2} . \cos\frac{x}{2}} = \frac{dt}{dx}\]
\[ \Rightarrow \frac{1}{\sin x} = \frac{dt}{dx}\]
\[ \Rightarrow \text{cosec x dx} = dt\]
\[ \therefore I = \int\frac{dt}{t}\]
\[ = \text{log}\left| t \right| + C\]
\[ = \text{log }\left| \log \tan\frac{x}{2} \right| + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\left( 2^x + \frac{5}{x} - \frac{1}{x^{1/3}} \right)dx\]
\[\int\frac{x^6 + 1}{x^2 + 1} dx\]
\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]
\[\int\sqrt{x}\left( 3 - 5x \right) dx\]
\[\int\frac{\cos^2 x - \sin^2 x}{\sqrt{1} + \cos 4x} dx\]
\[\int\frac{\left( x^3 + 8 \right)\left( x - 1 \right)}{x^2 - 2x + 4} dx\]
\[\int\left( x + 2 \right) \sqrt{3x + 5} \text{dx} \]
\[\int\frac{- \sin x + 2 \cos x}{2 \sin x + \cos x} dx\]
\[\int\frac{\cos^3 x}{\sqrt{\sin x}} dx\]
\[\int \tan^3 \text{2x sec 2x dx}\]
\[\int\frac{x + \sqrt{x + 1}}{x + 2} dx\]
\[\int\sqrt {e^x- 1} \text{dx}\]
\[\int\frac{x^2}{\sqrt{3x + 4}} dx\]
` ∫ tan^5 x dx `
\[\int \sin^3 x \cos^5 x \text{ dx }\]
\[\int\frac{x^4 + 1}{x^2 + 1} dx\]
\[\int\frac{x^2 + x + 1}{x^2 - x} dx\]
\[\int\frac{x + 1}{\sqrt{x^2 + 1}} dx\]
\[\int\frac{1}{4 \sin^2 x + 5 \cos^2 x} \text{ dx }\]
\[\int\frac{1}{13 + 3 \cos x + 4 \sin x} dx\]
\[\int\frac{1}{1 - \tan x} \text{ dx }\]
\[\int\text{ log }\left( x + 1 \right) \text{ dx }\]
\[\int x^2 \sin^2 x\ dx\]
\[\int e^\sqrt{x} \text{ dx }\]
\[\int e^x \frac{\left( 1 - x \right)^2}{\left( 1 + x^2 \right)^2} \text{ dx }\]
\[\int\left( x + 1 \right) \sqrt{x^2 + x + 1} \text{ dx }\]
\[\int\frac{5x}{\left( x + 1 \right) \left( x^2 - 4 \right)} dx\]
\[\int\frac{2x + 1}{\left( x - 2 \right) \left( x - 3 \right)} dx\]
\[\int\frac{1}{x^4 + 3 x^2 + 1} \text{ dx }\]
\[\int\frac{1}{\left( x - 1 \right) \sqrt{x^2 + 1}} \text{ dx }\]
\[\int\frac{x}{4 + x^4} \text{ dx }\] is equal to
\[\int \tan^3 x\ dx\]
\[\int \sin^3 x \cos^4 x\ \text{ dx }\]
\[\int\sqrt{\frac{1 + x}{x}} \text{ dx }\]
\[\int\sqrt{\frac{1 - x}{x}} \text{ dx}\]
\[\int\frac{x^3}{\sqrt{x^8 + 4}} \text{ dx }\]
\[\int\frac{x^2 + x + 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} \text{ dx}\]
\[\int\frac{\sin 4x - 2}{1 - \cos 4x} e^{2x} \text{ dx}\]
