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प्रश्न
\[\int\frac{\sin 2x}{\sin^4 x + \cos^4 x} \text{ dx }\]
योग
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उत्तर
\[\text{ Let I } = \int\frac{\sin 2x}{\sin^4 x + \cos^4 x}dx\]
\[ = \int\frac{2 \text{ sin x }\cdot \text{ cos x dx}}{\sin^4 x + \cos^4 x}\]
\[\text{Dividing numerator and denominator by} \cos^4 x\]
\[ \Rightarrow \int\frac{2 \frac{\text{ sin x }\cdot \text{ cos x}}{\cos^4 x}dx}{1 + \tan^4 x}\]
\[ \Rightarrow \int\frac{2 \tan x \cdot \text{ sec}^2 x dx}{1 + \left( \tan^2 x \right)^2}\]
\[\text{ Putting tan}^2 x = t\]
\[ \Rightarrow 2 \tan x \cdot \text{ sec}^2 \text{ x dx}\]
\[ \therefore I = \int\frac{dt}{1 + t^2}\]
\[ = \tan^{- 1} t + C\]
\[ = \tan^{- 1} \left( \text{ tan}^2 x \right) + C......... \left[ \because t = \tan {}^2 x \right]\]
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