Advertisements
Advertisements
प्रश्न
\[\int e^x \left( \cot x - {cosec}^2 x \right) dx\]
बेरीज
उत्तर
\[\text{ Let I }= \int e^x \left( \cot x - {cosec}^2 x \right)dx\]
\[\text{ here f(x) } = \text{ cot x put e}^x f(x) = t\]
\[ f'(x) = - {cosec}^2 x\]
\[\text{ let e}^x \cot x = t\]
\[\text{ Diff both sides w . r . t x }\]
\[ e^x \cot x + e^x \left( - {cosec}^2 x \right) = \frac{dt}{dx}\]
\[ \Rightarrow e^x \left( \cot x - {cosec}^2 x \right)dx = dt\]
\[ \therefore \int e^x \left( \cot x - {cosec}^2 x \right)dx = \int dt\]
\[ = t + C\]
\[ = e^x \cot x + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{x^2 + x + 5}{3x + 2} dx\]
\[\int\frac{3x + 5}{\sqrt{7x + 9}} dx\]
\[\int\frac{\text{sin} \left( x - a \right)}{\text{sin}\left( x - b \right)} dx\]
\[\int\frac{\cos^3 x}{\sqrt{\sin x}} dx\]
\[\int \tan^{3/2} x \sec^2 \text{x dx}\]
\[\int\frac{x \sin^{- 1} x^2}{\sqrt{1 - x^4}} dx\]
\[\int\frac{1}{x^2 \left( x^4 + 1 \right)^{3/4}} dx\]
\[\int x^2 \sqrt{x + 2} \text{ dx }\]
\[\int\frac{1}{\sqrt{x} + \sqrt[4]{x}}dx\]
\[\int \sin^5 x \cos x \text{ dx }\]
\[\int\frac{1}{\sin x \cos^3 x} dx\]
\[\int\frac{1}{\sqrt{1 + 4 x^2}} dx\]
\[\int\frac{1}{x \left( x^6 + 1 \right)} dx\]
\[\int\frac{x^2 + x + 1}{x^2 - x} dx\]
\[\int\frac{2x + 1}{\sqrt{x^2 + 2x - 1}}\text{ dx }\]
\[\int\frac{1}{4 \cos^2 x + 9 \sin^2 x}\text{ dx }\]
\[\int\frac{1}{3 + 2 \cos^2 x} \text{ dx }\]
\[\int\frac{\sin 2x}{\sin^4 x + \cos^4 x} \text{ dx }\]
`int 1/(sin x - sqrt3 cos x) dx`
\[\int x e^x \text{ dx }\]
\[\int \left( \log x \right)^2 \cdot x\ dx\]
\[\int {cosec}^3 x\ dx\]
\[\int\left( x + 1 \right) \text{ log x dx }\]
\[\int\sqrt{3 - 2x - 2 x^2} \text{ dx}\]
\[\int\frac{\sqrt{16 + \left( \log x \right)^2}}{x} \text{ dx}\]
\[\int\sqrt{x^2 - 2x} \text{ dx}\]
\[\int\frac{1}{x\left( x^n + 1 \right)} dx\]
\[\int\frac{1}{\sin x + \sin 2x} dx\]
\[\int\frac{x}{\left( x^2 + 2x + 2 \right) \sqrt{x + 1}} \text{ dx}\]
\[\int\frac{x}{\left( x^2 + 4 \right) \sqrt{x^2 + 1}} \text{ dx }\]
\[\int\frac{\sin^2 x}{\cos^4 x} dx =\]
\[\int\frac{x^3}{x + 1}dx\] is equal to
\[\int\frac{1}{\left( \sin^{- 1} x \right) \sqrt{1 - x^2}} \text{ dx} \]
\[\int\frac{1}{e^x + 1} \text{ dx }\]
\[\int\frac{1}{\sqrt{3 - 2x - x^2}} \text{ dx}\]
\[\int\frac{1}{1 - 2 \sin x} \text{ dx }\]
\[\int \sec^6 x\ dx\]
\[\int \tan^{- 1} \sqrt{x}\ dx\]
\[\int\frac{\cot x + \cot^3 x}{1 + \cot^3 x} \text{ dx}\]
\[\int\frac{x + 3}{\left( x + 4 \right)^2} e^x dx =\]
