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प्रश्न
\[\int\frac{1}{\sin^4 x + \cos^4 x} \text{ dx}\]
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उत्तर
\[\text{We have}, \]
\[I = \int\frac{dx}{\sin^4 x + \cos^4 x}\]
Dividing numerator and denominator by cos4x
\[I = \int\frac{\sec^4 \text{ x dx}}{\tan^4 x + 1}\]
\[ = \int\frac{\sec^2 x \sec^2 \text{ x dx}}{\tan^4 x + 1}\]
\[ = \int\frac{\left( 1 + \tan^2 x \right) \sec^2 \text{ x dx}}{\tan^4 x + 1}\]
\[\text{ Putting tan x = t}\]
\[ \Rightarrow \sec^2 \text{ x dx = dt}\]
\[ \therefore I = \int\frac{\left( 1 + t^2 \right) dt}{t^4 + 1}\]
\[ = \int\frac{\left( \frac{1}{t^2} + 1 \right) dt}{t^2 + \frac{1}{t^2}}\]
\[ = \int\frac{\left( 1 + \frac{1}{t^2} \right)}{\left( t - \frac{1}{t} \right)^2 + 2}dt\]
\[\text{ Putting t }- \frac{1}{t} = p\]
\[ \Rightarrow \left( 1 + \frac{1}{t^2} \right)dt = dp\]
\[ \therefore I = \int\frac{1}{p^2 + \left( \sqrt{2} \right)^2}dp\]
\[ = \frac{1}{\sqrt{2}} \text{ tan}^{- 1} \left( \frac{p}{\sqrt{2}} \right) + C\]
\[ = \frac{1}{\sqrt{2}} \text{ tan}^{- 1} \left( \frac{t - \frac{1}{t}}{\sqrt{2}} \right) + C\]
\[ = \frac{1}{\sqrt{2}} \text{ tan}^{- 1} \left( \frac{t^2 - 1}{\sqrt{2} \text{ t }} \right) + C\]
\[ = \frac{1}{\sqrt{2}} \text{ tan}^{- 1} \left( \frac{\tan^2 x - 1}{\sqrt{2} \tan x} \right) + C\]
\[ = \frac{1}{\sqrt{2}} \text{ tan}^{- 1} \left( - \sqrt{2} \times \frac{1 - \tan^2 x}{2 \tan x} \right) + C\]
\[ = \frac{1}{\sqrt{2}} \text{ tan}^{- 1} \left( \frac{- \sqrt{2}}{\tan 2x} \right) + C\]
\[ = \frac{1}{\sqrt{2}} \text{ tan}^{- 1} \left( - \sqrt{2} \cot 2x \right) + C\]
