Advertisements
Advertisements
प्रश्न
\[\int x\ {cosec}^2 \text{ x }\ \text{ dx }\]
योग
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
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{1}{1 + \cos 2x} dx\]
If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, f
\[\left( \frac{\pi}{2} \right)\] = 5, find f(x)
\[\int\frac{1}{\left( 7x - 5 \right)^3} + \frac{1}{\sqrt{5x - 4}} dx\]
\[\int\left( x + 2 \right) \sqrt{3x + 5} \text{dx} \]
\[\int\left( 5x + 3 \right) \sqrt{2x - 1} dx\]
` ∫ sin 4x cos 7x dx `
\[\int\text{sin mx }\text{cos nx dx m }\neq n\]
Integrate the following integrals:
\[\int\text { sin x cos 2x sin 3x dx}\]
` ∫ {"cosec" x }/ { log tan x/2 ` dx
\[\int\frac{1 - \sin 2x}{x + \cos^2 x} dx\]
\[\int \cos^5 x \text{ dx }\]
\[\int x \cos^3 x^2 \sin x^2 \text{ dx }\]
\[\int\frac{\sec^2 x}{1 - \tan^2 x} dx\]
\[\int\frac{1}{\sqrt{\left( 1 - x^2 \right)\left\{ 9 + \left( \sin^{- 1} x \right)^2 \right\}}} dx\]
\[\int\frac{\cos x - \sin x}{\sqrt{8 - \sin2x}}dx\]
\[\int\frac{x + 7}{3 x^2 + 25x + 28}\text{ dx}\]
\[\int\frac{x^2}{x^2 + 6x + 12} \text{ dx }\]
\[\int\frac{1}{4 \sin^2 x + 5 \cos^2 x} \text{ dx }\]
\[\int\frac{1}{2 + \sin x + \cos x} \text{ dx }\]
`int 1/(sin x - sqrt3 cos x) dx`
\[\int\frac{1}{5 + 7 \cos x + \sin x} dx\]
`int"x"^"n"."log" "x" "dx"`
\[\int e^x \frac{\left( 1 - x \right)^2}{\left( 1 + x^2 \right)^2} \text{ dx }\]
\[\int\frac{e^x}{x}\left\{ \text{ x }\left( \log x \right)^2 + 2 \log x \right\} dx\]
\[\int\left( 2x + 3 \right) \sqrt{x^2 + 4x + 3} \text{ dx }\]
\[\int\frac{5 x^2 - 1}{x \left( x - 1 \right) \left( x + 1 \right)} dx\]
\[\int\frac{\cos x}{\left( 1 - \sin x \right)^3 \left( 2 + \sin x \right)} dx\]
\[\int\frac{1}{x \left( x^4 - 1 \right)} dx\]
\[\int\frac{x^2 - 1}{x^4 + 1} \text{ dx }\]
\[\int\frac{x}{\left( x - 3 \right) \sqrt{x + 1}} dx\]
\[\int\frac{1}{\left( 1 + x^2 \right) \sqrt{1 - x^2}} \text{ dx }\]
\[\int x^{\sin x} \left( \frac{\sin x}{x} + \cos x . \log x \right) dx\] is equal to
\[\int\left( x - 1 \right) e^{- x} dx\] is equal to
\[\int \tan^4 x\ dx\]
\[\int\sqrt{\text{ cosec x} - 1} \text{ dx }\]
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
\[\int\sqrt{\frac{a + x}{x}}dx\]
\[\int\frac{1}{\sec x + cosec x}\text{ dx }\]
\[\int x\sqrt{1 + x - x^2}\text{ dx }\]
\[\int \left( e^x + 1 \right)^2 e^x dx\]
