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∫ 1 √ 1 + Cos X D X - Mathematics

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प्रश्न

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

उत्तर

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

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

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

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

`= \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 `

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अध्याय 19: Indefinite Integrals - Exercise 19.08 [पृष्ठ ४७]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.08 | Q 2 | पृष्ठ ४७

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