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∫ Sin 2 X √ Sin 4 X + 4 Sin 2 X − 2 D X - Mathematics

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Question

\[\int\frac{\sin 2x}{\sqrt{\sin^4 x + 4 \sin^2 x - 2}} dx\]
Sum
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Solution

` ∫   {  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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Chapter 19: Indefinite Integrals - Exercise 19.18 [Page 99]

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RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.18 | Q 10 | Page 99

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