Advertisements
Advertisements
प्रश्न
\[\int x\ {cosec}^2 \text{ x }\ \text{ dx }\]
बेरीज
Advertisements
उत्तर
\[\int x\ {cosec}^2 \text{ x }\ \text{ dx }\]
` " Taking x as the first function and cosec"^2 x " as the second function " . `
\[ = x\int {cosec}^2 x\ dx - \int\left\{ \frac{d}{dx}\left( x \right)\int {cosec}^2 x\ dx \right\}dx\]
\[ = - x \text{ cot x } + \int\text{ cot x dx }\]
\[ = - x \text{ cot x }+ \text{ log }\left| \sin x \right| + c\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
`int{sqrtx(ax^2+bx+c)}dx`
\[\int\sqrt{x}\left( x^3 - \frac{2}{x} \right) dx\]
\[\int \cos^{- 1} \left( \sin x \right) dx\]
\[\int\frac{\cos x}{1 + \cos x} dx\]
\[\int\left( x + 2 \right) \sqrt{3x + 5} \text{dx} \]
` ∫ sin x \sqrt (1-cos 2x) dx `
\[\int\frac{\sec^2 x}{\tan x + 2} dx\]
\[\int\frac{\log\left( 1 + \frac{1}{x} \right)}{x \left( 1 + x \right)} dx\]
\[\int\frac{\sin\sqrt{x}}{\sqrt{x}} dx\]
\[\int\frac{e^{m \tan^{- 1} x}}{1 + x^2} dx\]
\[\int\frac{x + \sqrt{x + 1}}{x + 2} dx\]
\[\int\sqrt {e^x- 1} \text{dx}\]
\[\int\left( 2 x^2 + 3 \right) \sqrt{x + 2} \text{ dx }\]
` ∫ 1 /{x^{1/3} ( x^{1/3} -1)} ` dx
\[\int \cot^n {cosec}^2 \text{ x dx } , n \neq - 1\]
\[\int \cos^7 x \text{ dx } \]
\[\int\frac{x^2 - 1}{x^2 + 4} dx\]
\[\int\frac{1}{\sqrt{\left( 1 - x^2 \right)\left\{ 9 + \left( \sin^{- 1} x \right)^2 \right\}}} dx\]
\[\int\frac{\sin x - \cos x}{\sqrt{\sin 2x}} dx\]
\[\int\frac{a x^3 + bx}{x^4 + c^2} dx\]
\[\int\frac{x^2 + 1}{x^2 - 5x + 6} dx\]
\[\int\frac{x^2}{x^2 + 7x + 10} dx\]
\[\int\frac{3x + 1}{\sqrt{5 - 2x - x^2}} \text{ dx }\]
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
`int 1/(sin x - sqrt3 cos x) dx`
\[\int\left( 4x + 1 \right) \sqrt{x^2 - x - 2} \text{ dx }\]
\[\int\left( 2x + 3 \right) \sqrt{x^2 + 4x + 3} \text{ dx }\]
\[\int(2x + 5)\sqrt{10 - 4x - 3 x^2}dx\]
\[\int\frac{dx}{\left( x^2 + 1 \right) \left( x^2 + 4 \right)}\]
\[\int\frac{1}{\left( x - 1 \right) \sqrt{x + 2}} \text{ dx }\]
\[\int\frac{1}{\left( x - 1 \right) \sqrt{2x + 3}} \text{ dx }\]
The value of \[\int\frac{\sin x + \cos x}{\sqrt{1 - \sin 2x}} dx\] is equal to
\[\int\frac{\left( 2^x + 3^x \right)^2}{6^x} \text{ dx }\]
\[\int \cot^5 x\ dx\]
\[\int \sin^3 x \cos^4 x\ \text{ dx }\]
\[\int\frac{x + 1}{x^2 + 4x + 5} \text{ dx}\]
\[\int \tan^5 x\ \sec^3 x\ dx\]
