हिंदी

∫ Cot N C O S E C 2 X D X , N ≠ − 1 - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int \cot^n {cosec}^2 \text{ x dx } , n \neq - 1\]
योग
Advertisements

उत्तर

∫ cotn x cosec2 x dx
Let cot x = t
⇒ –cosec2 x dx = dt
⇒ cosec2 x dx = –dt

\[Now, \int \cot^n \text{ x } {cosec}^2  \text  { x dx  }\]
\[ = - \int t^n dt \]
\[ = \frac{- t^{n + 1}}{n + 1} + C\]
\[ = - \frac{\cot^{n + 1} x}{n + 1} + C\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Exercise 19.11 [पृष्ठ ६९]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.11 | Q 9 | पृष्ठ ६९

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

\[\int\frac{x^6 + 1}{x^2 + 1} dx\]

\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]

\[\int\sqrt{x}\left( 3 - 5x \right) dx\]

 


\[\int\frac{1 - \cos x}{1 + \cos x} dx\]

\[\int\frac{x^3}{x - 2} dx\]

\[\int     \text{sin}^2  \left( 2x + 5 \right)    \text{dx}\]

Integrate the following integrals:

\[\int\text{sin 2x  sin 4x    sin 6x  dx} \]

\[\int\frac{1}{      x      \text{log x } \text{log }\left( \text{log x }\right)} dx\]

\[\int\frac{1}{1 + \sqrt{x}} dx\]

\[\int\frac{\sin^5 x}{\cos^4 x} \text{ dx }\]

\[\int x^2 \sqrt{x + 2} \text{  dx  }\]

\[\int\frac{1}{\sin x \cos^3 x} dx\]

Evaluate the following integrals:
\[\int\frac{x^2}{\left( a^2 - x^2 \right)^{3/2}}dx\]

\[\int\frac{1}{\sqrt{a^2 + b^2 x^2}} dx\]

\[\int\frac{1}{x^2 + 6x + 13} dx\]

\[\int\frac{dx}{e^x + e^{- x}}\]

\[\int\frac{1}{\sqrt{7 - 6x - x^2}} dx\]

\[\int\frac{x}{\sqrt{8 + x - x^2}} dx\]


\[\int\frac{1}{\sin x + \sqrt{3} \cos x} \text{ dx  }\]

\[\int x \cos^2 x\ dx\]

`int"x"^"n"."log"  "x"  "dx"`

\[\int x\left( \frac{\sec 2x - 1}{\sec 2x + 1} \right) dx\]

\[\int x \sin^3 x\ dx\]

\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2}  \text{ dx }\]

\[\int\left( 4x + 1 \right) \sqrt{x^2 - x - 2} \text{  dx }\]

\[\int\left( x - 2 \right) \sqrt{2 x^2 - 6x + 5} \text{  dx }\]

\[\int\left( 2x - 5 \right) \sqrt{x^2 - 4x + 3} \text{  dx }\]

 


\[\int\frac{x^3 - 1}{x^3 + x} dx\]

\[\int\frac{1}{\cos x \left( 5 - 4 \sin x \right)} dx\]

\[\int\frac{x}{\left( x^2 + 2x + 2 \right) \sqrt{x + 1}} \text{ dx}\]

\[\int\frac{x^3}{x + 1}dx\] is equal to

\[\int \text{cosec}^2 x \text{ cos}^2 \text{  2x  dx} \]

\[\int\frac{x^3}{\left( 1 + x^2 \right)^2} \text{ dx }\]

\[\int \tan^5 x\ \sec^3 x\ dx\]

\[ \int\left( 1 + x^2 \right) \ \cos 2x \ dx\]


\[\int\frac{x^2}{\sqrt{1 - x}} \text{ dx }\]

\[\int\frac{x^2}{\left( x - 1 \right)^3 \left( x + 1 \right)} \text{ dx}\]

\[\int\frac{x}{x^3 - 1} \text{ dx}\]

Evaluate : \[\int\frac{\cos 2x + 2 \sin^2 x}{\cos^2 x}dx\] .


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×