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प्रश्न
\[\int\frac{\sin 2x}{\sqrt{\sin^4 x + 4 \sin^2 x - 2}} dx\]
बेरीज
उत्तर
` ∫ { sin (2 x ) dx }/{\sqrt{ sin^4 x + 4 sin^2 x-2}} `
`\text{ let }\sin^2 x = t `
` ⇒ 2 sin x cos x dx = dt`
\[ \Rightarrow \text{ sin }\left( 2 x \right) dx = dt\]
Now, ` ∫ { sin (2 x ) dx }/{\sqrt{ sin^4 x + 4 sin^2 x-2}} `
\[ = \int\frac{dt}{\sqrt{t^2 + 4t - 2}}\]
\[ = \int\frac{dt}{\sqrt{t^2 + 4t + 4 - 4 - 2}}\]
\[ = \int\frac{dt}{\sqrt{\left( t + 2 \right)^2 - \left( \sqrt{6} \right)^2}}\]
\[ = \text{ log }\left| t + 2 + \sqrt{\left( t + 2 \right)^2 - 6} \right| + C\]
\[ = \text{ log }\left| t + 2 + \sqrt{t^2 + 4t - 2} \right| + C\]
` = \text{ log } |sin^2 x + 2 + \sqrt{\sin^4 x + 4 \sin^2 x - 2} | + C`
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