English

∫ 1 1 − Sin X 2 D X - Mathematics

Advertisements
Advertisements

Question

\[\int\frac{1}{1 - \sin\frac{x}{2}} dx\]
Sum
Advertisements

Solution

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

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

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

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

\[ = \int\left( \sec^2 \frac{x}{2} + \sec \frac{x}{2} \text{tan }\frac{x}{2} \right)dx\]

\[ = \frac{\tan \left( \frac{x}{2} \right)}{\frac{1}{2}} + \frac{\sec \left( \frac{x}{2} \right)}{\frac{1}{2}} + C\]

\[ = 2 \left( \tan \frac{x}{2} + \sec \frac{x}{2} \right) + C\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Exercise 19.03 [Page 23]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.03 | Q 12 | Page 23

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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

\[\int\left( x^e + e^x + e^e \right) dx\]

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

\[\int\frac{1 - \cos 2x}{1 + \cos 2x} dx\]

` ∫  {cosec x} / {"cosec x "- cot x} ` dx      


\[\int\frac{1}{\left( 7x - 5 \right)^3} + \frac{1}{\sqrt{5x - 4}} dx\]

\[\int\frac{x}{\sqrt{x + a} - \sqrt{x + b}}dx\]

\[\int \cos^2 \text{nx dx}\]

\[\int\frac{\cos x - \sin x}{1 + \sin 2x} dx\]

\[\int x^3 \cos x^4 dx\]

\[\int 5^{x + \tan^{- 1} x} . \left( \frac{x^2 + 2}{x^2 + 1} \right) dx\]

\[\  ∫    x   \text{ e}^{x^2} dx\]

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

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

\[\int\frac{1}{\sin^4 x \cos^2 x} dx\]

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

\[\int\frac{1}{\sqrt{1 + 4 x^2}} dx\]

 


\[\int\frac{x}{x^4 + 2 x^2 + 3} dx\]

\[\int\frac{1}{5 + 7 \cos x + \sin x} dx\]

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

\[\int \sin^{- 1} \left( 3x - 4 x^3 \right) \text{ dx }\]

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

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

\[\int\frac{x^2 + 6x - 8}{x^3 - 4x} dx\]

If \[\int\frac{1}{5 + 4 \sin x} dx = A \tan^{- 1} \left( B \tan\frac{x}{2} + \frac{4}{3} \right) + C,\] then


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

\[\int\frac{x^9}{\left( 4 x^2 + 1 \right)^6}dx\]  is equal to 

\[\int\frac{1 - x^4}{1 - x} \text{ dx }\]


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

\[\int\frac{\sin x - \cos x}{\sqrt{\sin 2x}} \text{ dx }\]
 
 

\[\int\frac{1}{\text{ sin} \left( x - a \right) \text{ sin } \left( x - b \right)} \text{ dx }\]

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

\[\int\sqrt{\text{ cosec  x} - 1} \text{ dx }\]

\[\int\frac{1}{\sqrt{3 - 2x - x^2}} \text{ dx}\]

\[\int\frac{1}{\sin x + \sin 2x} \text{ dx }\]

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

\[\int\frac{x^5}{\sqrt{1 + x^3}} \text{ dx }\]

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

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


Find: `int (3x +5)/(x^2+3x-18)dx.`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×