English
CBSECommerce (English Medium) Class 12
Question Papers2580
Textbook Solutions20345
MCQ Online Mock Tests42
Important Solutions22654
Concept Notes & Videos608
Time Tables24
Syllabus

∫ 1 1 + Cos 3 X D X - Mathematics

Advertisements
Advertisements

Question

` ∫  1/ {1+ cos   3x}  ` dx

Sum
Advertisements

Solution

` ∫  1/ {1+ cos   3x}  ` dx

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

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

\[ = \int\left( \frac{1 - \cos 3x}{\sin^2 3x} \right) dx\]

\[ = \int \text{cosec}^\text{2}\text{ 3x dx} -  ∫cosec\ 3x \cot 3xdx\]

` = - {cot 3x} / 3 + {"cosec "  3x} / 3 + c `

` = 1/3 [ "cosec"   3x - cot 3x ] + c ` 

\[ = \frac{1}{3}\left[ \frac{1}{\sin 3x} - \frac{\cos 3x}{\sin 3x} \right] + C\]

\[ = \frac{1}{3} \left[ \frac{1 - \cos 3x}{\sin 3x} \right] + C\]

shaalaa.com
Indefinite Integral Problems
  Is there an error in this question or solution?
Q 13
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 13 | Page 23

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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

\[\int\frac{5 \cos^3 x + 6 \sin^3 x}{2 \sin^2 x \cos^2 x} dx\]

\[\int\frac{\left( x^3 + 8 \right)\left( x - 1 \right)}{x^2 - 2x + 4} dx\]

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

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

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

` ∫   sin x  \sqrt (1-cos 2x)    dx `

 


`  ∫  sin 4x cos  7x  dx  `

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

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

` ∫    \sqrt{tan x}     sec^4  x   dx `


\[\int \sec^4 2x \text{ dx }\]

\[\int \cot^6 x \text{ dx }\]

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

\[\int \sin^3 x \cos^6 x \text{ dx }\]

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

\[\int\frac{x}{\sqrt{x^2 + 6x + 10}} \text{ dx }\]

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

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

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

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

\[\int x \sin x \cos x\ dx\]

 


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

\[\int\left( x + 1 \right) \text{ e}^x \text{ log } \left( x e^x \right) dx\]

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

\[\int \sin^3 \sqrt{x}\ dx\]

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

\[\int\frac{\sqrt{16 + \left( \log x \right)^2}}{x} \text{ dx}\]

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

 


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

\[\int\frac{1}{x \left( x^4 + 1 \right)} dx\]

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

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

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

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


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

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

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

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


Share
Notifications

Englishहिंदीमराठी
Login
Create free account


      Forgot password?
Course
Use app×