Question

Evaluate the following integral:

\[\int\frac{1}{\sin^4 x + \sin^2 x \cos^2 x + \cos^4 x}dx\]

Answer 1

\[\text{ Let I} = \int\frac{1}{\sin^4 x + \sin^2 x \cos^2 x + \cos^4 x}dx\]
\[ = \int\frac{1}{\left( \sin^2 x + \cos^2 x \right)^2 - \sin^2 x \cos^2 x}dx\]
\[ = \int\frac{1}{1 - \sin^2 x \cos^2 x}dx\]
\[ = \int\frac{\frac{1}{\cos^4 x}}{\frac{1}{\cos^4 x} - \frac{\sin^2 x}{\cos^2 x}}dx\]
\[ = \int\frac{\sec^2 x\left( 1 + \tan^2 x \right)}{\sec^4 x - \tan^2 x}dx\]
\[ \text{ Let  tan x }= t\]
\[ \text{On differentiating both sides, we get}\]
\[ \sec^2 \text{ x dx }= dt\]
\[ \therefore I = \int\frac{1 + t^2}{\left( 1 + t^2 \right)^2 - t^2}dt\]
\[ = \int\frac{1 + t^2}{\left( t^4 + t^2 + 1 \right)}dt\]
\[ = \int\frac{\frac{1}{t^2} + 1}{\left( t^2 + 1 + \frac{1}{t^2} \right)}dt\]
\[ = \int\frac{\frac{1}{t^2} + 1}{\left( t - \frac{1}{t} \right)^2 + 3}dt\]
\[ \text{ Let }\left( t - \frac{1}{t} \right) = u\]
\[ \text{On differentiating both sides, we get}\]
\[ \left( 1 + \frac{1}{t^2} \right) dt = du\]
\[ \therefore I = \int\frac{1}{\left( u \right)^2 + 3}du\]
\[ = \frac{1}{\sqrt{3}} \tan^{- 1} \left( \frac{u}{\sqrt{3}} \right) + c\]
\[ = \frac{1}{\sqrt{3}} \tan^{- 1} \left( \frac{t - \frac{1}{t}}{\sqrt{3}} \right) + c\]
\[ = \frac{1}{\sqrt{3}} \tan^{- 1} \left( \frac{\tan x - \cot x}{\sqrt{3}} \right) + c\]
\[\text{ Hence,} \int\frac{1}{\sin^4 x + \sin^2 x \cos^2 x + \cos^4 x}dx = \frac{1}{\sqrt{3}} \tan^{- 1} \left( \frac{\tan x - \cot x}{\sqrt{3}} \right) + c\]