Advertisements
Advertisements
प्रश्न
\[\int \cot^4 x\ dx\]
बेरीज
Advertisements
उत्तर
\[\text{ Let I }= \int \cot^4 \text{ x dx}\]
\[ = \int \cot^2 x \cdot \cot^2 \text{ x dx}\]
\[ = \int \cot^2 x \cdot \left( \text{ cosec}^2 x - 1 \right) \text{ dx}\]
\[ = \int \cot^2 x \cdot \text{ cosec }^2 \text{ x dx} - \int \cot^2 \text{ x dx}\]
\[ = \int \cot^2 x \cdot\text {cosec}^2 \text{ x dx}- \int\left( \text{cosec}^2 x - 1 \right) \text{ dx}\]
\[ \text{ Putting cot x = t in the Ist integral}\]
\[ \Rightarrow - \text{ cosec}^2 \text{ x dx} = dt\]
\[ \therefore I = - \int t^2 dt - \int\left( \text{cosec}^2 x - 1 \right) \text{ dx}\]
\[ = \frac{- t^3}{3} + \text{ cot x + x + C}\]
\[ = \frac{- \cot^3 x}{3} + \text{ cot x + x + C}................\left[ \because t = \text{ cot x} \right]\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\left( \frac{m}{x} + \frac{x}{m} + m^x + x^m + mx \right) dx\]
\[\int\left( \sec^2 x + {cosec}^2 x \right) dx\]
\[\int\frac{1 - \cos x}{1 + \cos x} dx\]
\[\int\frac{1}{\text{cos}^2\text{ x }\left( 1 - \text{tan x} \right)^2} dx\]
\[\int\frac{\text{sin} \left( x - \alpha \right)}{\text{sin }\left( x + \alpha \right)} dx\]
\[\int\frac{\sin 2x}{\sin 5x \sin 3x} dx\]
\[\int\frac{\left( x + 1 \right) e^x}{\sin^2 \left( \text{x e}^x \right)} dx\]
\[\ ∫ x \text{ e}^{x^2} dx\]
\[\int\sqrt {e^x- 1} \text{dx}\]
\[\int\frac{x^2}{\sqrt{3x + 4}} dx\]
\[\int \sin^3 x \cos^5 x \text{ dx }\]
\[\int\frac{1}{\sin^3 x \cos^5 x} dx\]
\[\int\frac{1}{2 x^2 - x - 1} dx\]
\[\int\frac{x^2 + x + 1}{x^2 - x + 1} \text{ dx }\]
\[\int\frac{x + 1}{\sqrt{4 + 5x - x^2}} \text{ dx }\]
\[\int\sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]
\[\int\frac{1}{4 \sin^2 x + 5 \cos^2 x} \text{ dx }\]
\[\int\frac{1}{5 + 4 \cos x} dx\]
\[\int\frac{1}{13 + 3 \cos x + 4 \sin x} dx\]
\[\int\frac{1}{3 + 4 \cot x} dx\]
\[\int\frac{2 \tan x + 3}{3 \tan x + 4} \text{ dx }\]
\[\int x^2 \text{ cos x dx }\]
\[\int \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) \text{ dx }\]
\[\int x^2 \tan^{- 1} x\text{ dx }\]
\[\int x \sin^3 x\ dx\]
\[\int e^x \left( \frac{x - 1}{2 x^2} \right) dx\]
\[\int e^x \left( \frac{\sin 4x - 4}{1 - \cos 4x} \right) dx\]
\[\int\left\{ \tan \left( \log x \right) + \sec^2 \left( \log x \right) \right\} dx\]
\[\int\left( x + 2 \right) \sqrt{x^2 + x + 1} \text{ dx }\]
\[\int\left( 2x - 5 \right) \sqrt{x^2 - 4x + 3} \text{ dx }\]
\[\int\frac{x^2 + 1}{x^2 - 1} dx\]
\[\int\frac{dx}{\left( x^2 + 1 \right) \left( x^2 + 4 \right)}\]
\[\int\frac{x + 1}{x \left( 1 + x e^x \right)} dx\]
\[\int\frac{\left( x - 1 \right)^2}{x^4 + x^2 + 1} \text{ dx}\]
\[\int\frac{\cos2x - \cos2\theta}{\cos x - \cos\theta}dx\] is equal to
\[\int\frac{1}{\left( \sin^{- 1} x \right) \sqrt{1 - x^2}} \text{ dx} \]
\[\int\frac{e^x - 1}{e^x + 1} \text{ dx}\]
\[\int\frac{\sin x}{\cos 2x} \text{ dx }\]
\[\int\frac{1}{\left( \sin x - 2 \cos x \right) \left( 2 \sin x + \cos x \right)} \text{ dx }\]
\[\int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]
