Advertisements
Advertisements
प्रश्न
\[\int\frac{1 - \cot x}{1 + \cot x} dx\]
बेरीज
उत्तर
\[\text{Let I} = \int\left( \frac{1 - \cot x}{1 + \cot x} \right)dx\]
\[ = \int\left( \frac{1 - \frac{\cos x}{\sin x}}{1 + \frac{\cos x}{\sin x}} \right)dx\]
\[ = \int\left( \frac{\sin x - \cos x}{\sin x + \cos x} \right)dx\]
\[\text{Putting }\sin x + \cos x = t\]
\[ \Rightarrow \left( \cos x - \sin x \right)dx = dt\]
\[ \Rightarrow \left( \sin x - \cos x \right)dx = - dt\]
\[ \therefore I = \int\frac{- dt}{t}\]
\[ = - \text{ln }\left| t \right| + C\]
\[ = - \text{ln} \left| \sin x + \cos x \right| + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\left( 3x\sqrt{x} + 4\sqrt{x} + 5 \right)dx\]
\[\int\frac{1}{\sqrt{x}}\left( 1 + \frac{1}{x} \right) dx\]
\[\int\frac{1 - \cos x}{1 + \cos x} dx\]
\[\int\frac{x^2 + 3x - 1}{\left( x + 1 \right)^2} dx\]
` ∫ cos mx cos nx dx `
` ∫ {sec x "cosec " x}/{log ( tan x) }` dx
\[\int\frac{- \sin x + 2 \cos x}{2 \sin x + \cos x} dx\]
\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]
\[\int2x \sec^3 \left( x^2 + 3 \right) \tan \left( x^2 + 3 \right) dx\]
\[\int\sqrt {e^x- 1} \text{dx}\]
\[\int\frac{1}{\sqrt{x} + \sqrt[4]{x}}dx\]
\[\int \sec^4 2x \text{ dx }\]
\[\int {cosec}^4 \text{ 3x } \text{ dx } \]
\[\int\frac{x}{3 x^4 - 18 x^2 + 11} dx\]
\[\int\frac{1}{\sqrt{3 x^2 + 5x + 7}} dx\]
\[\int\frac{\sin 2x}{\sqrt{\sin^4 x + 4 \sin^2 x - 2}} dx\]
\[\int\frac{\sin 2x}{\sqrt{\cos^4 x - \sin^2 x + 2}} dx\]
\[\int\frac{2x + 1}{\sqrt{x^2 + 2x - 1}}\text{ dx }\]
\[\int\frac{1}{\cos x \left( \sin x + 2 \cos x \right)} dx\]
\[\int\frac{1}{\sqrt{3} \sin x + \cos x} dx\]
\[\int\frac{1}{p + q \tan x} \text{ dx }\]
\[\int e^x \left( \tan x - \log \cos x \right) dx\]
\[\int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx\]
\[\int e^x \sec x \left( 1 + \tan x \right) dx\]
\[\int e^x \frac{\left( 1 - x \right)^2}{\left( 1 + x^2 \right)^2} \text{ dx }\]
\[\int\sqrt{2ax - x^2} \text{ dx}\]
\[\int\left( 2x - 5 \right) \sqrt{x^2 - 4x + 3} \text{ dx }\]
\[\int\frac{1}{\left( x - 1 \right) \left( x + 1 \right) \left( x + 2 \right)} dx\]
\[\int\frac{3x + 5}{x^3 - x^2 - x + 1} dx\]
\[\int\frac{1}{\sin x \left( 3 + 2 \cos x \right)} dx\]
\[\int\frac{\left( x^2 + 1 \right) \left( x^2 + 2 \right)}{\left( x^2 + 3 \right) \left( x^2 + 4 \right)} dx\]
\[\int\frac{\sin^6 x}{\cos^8 x} dx =\]
\[\int\frac{\sin x + \cos x}{\sqrt{\sin 2x}} \text{ dx}\]
\[\int\frac{1}{x^2 + 4x - 5} \text{ dx }\]
\[\int\frac{1}{1 + 2 \cos x} \text{ dx }\]
\[\int\sqrt{1 + 2x - 3 x^2}\text{ dx } \]
Find : \[\int\frac{dx}{\sqrt{3 - 2x - x^2}}\] .
Find : \[\int\frac{e^x}{\left( 2 + e^x \right)\left( 4 + e^{2x} \right)}dx.\]
\[\int \left( e^x + 1 \right)^2 e^x dx\]
