English

∫ 1 √ 1 + Cos X D X - Mathematics

Advertisements
Advertisements

Question

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

Solution

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

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

\[ = \frac{1}{\sqrt{2}}\int\sec\frac{x}{2}\text{ dx}\]

\[ = \frac{1}{\sqrt{2}} \times \text{2 }\text{ln }\left| \tan\frac{x}{2} + \sec\frac{x}{2} \right| + C\]

`= \sqrt2  In  |  {1 + sin ^ {x/2 }}/{cos ^{x/2}} | + C `

`= \sqrt2  In  | (( \text{sin} x/4 + \text{cos} x/4)^2  )/((cos^2  x /4  - \text{sin}^2 x /4 ))  | + C ` ` [ ∵  1 + sin θ  = ( sin^2  θ/2  + cos^2   θ/2+ 2 sin  θ/2 cos  θ/2 ) = ( sin  θ/2 + cos  θ/2)^2 and cos   θ = cos ^2  θ/2   - sin^2  θ/2 ]` 

`= \sqrt2  In  | (( \text{sin} x/4 + \text{cos} x/4)^2  )/((\text{cos }x /4  - \text{sin} x /4 )  (\text{cos} x/4  + \text{sin}x/4 ))| + C `

 

 

`= \sqrt2  In  | { sin  x/4 + cos  x/4} / {cos  x/4  - sin  x/4} |` + C 

`= \sqrt2  In  |  {1 + tan  x/4 } /{1- tan  x/4}| + C `

`= \sqrt2  In  |   tan  (x/4 + x/4) |+ C `

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Exercise 19.08 [Page 47]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.08 | Q 2 | Page 47

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

`int{sqrtx(ax^2+bx+c)}dx`

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

\[\int \tan^2 \left( 2x - 3 \right) dx\]


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

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

` ∫   e^{m   sin ^-1  x}/ \sqrt{1-x^2}  ` dx

 


\[\int\sqrt {e^x- 1}  \text{dx}\] 

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


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

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

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

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

\[\int\frac{x + 2}{2 x^2 + 6x + 5}\text{  dx }\]

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

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

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

\[\int\frac{1}{4 + 3 \tan x} dx\]

\[\int\frac{x^3 \sin^{- 1} x^2}{\sqrt{1 - x^4}} \text{ dx }\]

\[\int e^x \left( \tan x - \log \cos x \right) dx\]

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

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

\[\int\frac{18}{\left( x + 2 \right) \left( x^2 + 4 \right)} dx\]

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

Evaluate the following integral:

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

\[\int\frac{x^2 + 9}{x^4 + 81} \text{ dx }\]

 


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

\[\int\frac{1}{x^4 + 3 x^2 + 1} \text{ dx }\]

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


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

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

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

\[\int\frac{1}{3 x^2 + 13x - 10} \text{ dx }\]

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

\[\int\frac{1}{4 \sin^2 x + 4 \sin x \cos x + 5 \cos^2 x} \text{ dx }\]


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


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

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

\[\int\frac{\cos^7 x}{\sin x} dx\]

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×