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प्रश्न
\[\int\frac{1}{\sqrt{1 - \cos 2x}} dx\]
योग
उत्तर
\[\int\frac{1}{\sqrt{1 - \cos 2x}}dx\]
\[ = \int\frac{1}{\sqrt{2 \sin^2 x}}dx \left[ \because 1 - \cos 2x = 2 \ sin^2 x \right]\]
` = 1/sqrt2 ∫ "cosec" x dx `
\[ = \frac{1}{\sqrt{2}}\text{ln }\left| \text{cosec x} - \text{ cot x} \right| + C\]
` = 1/\sqrt{2} In | 1/ sin x - cos x / sin x| + C`
` = 1/\sqrt{2} In | {2 sin ^{2 x/2}} / sin x | + C ` ` [ ∵ 1 - cos x = 2 sin^2 x/2 ]`
` = 1/\sqrt{2} In | {2 sin ^{2x/2}} / {2sin x/2 cos x/2 } |` + C ` [ ∵ sin x = 2 sin x/2 cos x/2 ]`
\[ = \frac{1}{\sqrt{2}} \text{ln} \left| \tan\frac{x}{2} \right| + C\]
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