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प्रश्न
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 `
