Advertisements
Advertisements
प्रश्न
\[\int\frac{\left( x + 1 \right) e^x}{\sin^2 \left( \text{x e}^x \right)} dx\]
बेरीज
उत्तर
\[\int\frac{\left( x + 1 \right) e^x}{\sin^2 \left( \text{x e}^x \right)}dx\]
\[\text{Let x e}^x = t\]
\[ \Rightarrow \left( 1 \cdot e^x + \text{x e}^x \right) = \frac{dt}{dx}\]
\[ \Rightarrow \left( x + 1 \right) \text{e}^x dx = dt\]
\[Now, \int\frac{\left( x + 1 \right) e^x}{\sin^2 \left( \text{x e}^x \right)}dx\]
\[ = \int\frac{dt}{\sin^2 t}\]
\[ = \int {cosec}^2 \text{t dt} \]
\[ = - \text{cot} \left( t \right) + C\]
\[ = - \text{cot} \left( \text{x e}^x \right) + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{2 x^4 + 7 x^3 + 6 x^2}{x^2 + 2x} dx\]
\[\int \left( e^x + 1 \right)^2 e^x dx\]
`∫ cos ^4 2x dx `
\[\int\frac{\cos x}{2 + 3 \sin x} dx\]
\[\int\frac{1}{\sqrt{1 - x^2}\left( 2 + 3 \sin^{- 1} x \right)} dx\]
` = ∫ root (3){ cos^2 x} sin x dx `
\[\int\frac{x^2}{\sqrt{1 - x}} dx\]
\[\int {cosec}^4 \text{ 3x } \text{ dx } \]
\[\int \cot^6 x \text{ dx }\]
\[\int \sin^3 x \cos^6 x \text{ dx }\]
\[\int\frac{1}{\sin^3 x \cos^5 x} dx\]
\[\int\frac{1}{\sqrt{a^2 + b^2 x^2}} dx\]
\[\int\frac{x^2}{x^6 + a^6} dx\]
` ∫ \sqrt{"cosec x"- 1} dx `
\[\int\frac{\sin x - \cos x}{\sqrt{\sin 2x}} dx\]
\[\int\frac{x + 2}{\sqrt{x^2 - 1}} \text{ dx }\]
\[\int\frac{x - 1}{\sqrt{x^2 + 1}} \text{ dx }\]
\[\int\frac{2}{2 + \sin 2x}\text{ dx }\]
\[\int\frac{1}{4 \cos x - 1} \text{ dx }\]
`int 1/(cos x - sin x)dx`
\[\int x^2 \text{ cos x dx }\]
` ∫ sin x log (\text{ cos x ) } dx `
\[\int \cos^{- 1} \left( 4 x^3 - 3x \right) \text{ dx }\]
\[\int\frac{x \sin^{- 1} x}{\sqrt{1 - x^2}} dx\]
\[\int e^x \left( \frac{x - 1}{2 x^2} \right) dx\]
\[\int e^x \left[ \sec x + \log \left( \sec x + \tan x \right) \right] dx\]
\[\int e^x \cdot \frac{\sqrt{1 - x^2} \sin^{- 1} x + 1}{\sqrt{1 - x^2}} \text{ dx }\]
\[\int\left( \frac{1}{\log x} - \frac{1}{\left( \log x \right)^2} \right) dx\]
\[\int\left\{ \tan \left( \log x \right) + \sec^2 \left( \log x \right) \right\} dx\]
\[\int\frac{x^2 + 1}{\left( 2x + 1 \right) \left( x^2 - 1 \right)} dx\]
\[\int\frac{\left( x^2 + 1 \right) \left( x^2 + 2 \right)}{\left( x^2 + 3 \right) \left( x^2 + 4 \right)} dx\]
\[\int\frac{4 x^4 + 3}{\left( x^2 + 2 \right) \left( x^2 + 3 \right) \left( x^2 + 4 \right)} dx\]
\[\int\frac{\cos 2x - 1}{\cos 2x + 1} dx =\]
\[\int \tan^3 x\ dx\]
\[\int \cot^5 x\ dx\]
\[\int\frac{1}{\sin x \left( 2 + 3 \cos x \right)} \text{ dx }\]
\[\int\frac{1}{2 + \cos x} \text{ dx }\]
\[\int\frac{x^5}{\sqrt{1 + x^3}} \text{ dx }\]
\[\int x^2 \tan^{- 1} x\ dx\]