Question

\[\int_0^\frac{\pi}{4} \frac{\sin^2 x \cos^2 x}{\left( \sin^3 x + \cos^3 x \right)^2}dx\]

Answer 1

Let

\[I = \int_0^\frac{\pi}{4} \frac{\sin^2 x \cos^2 x}{\left( \sin^3 x + \cos^3 x \right)^2}dx\]

\[= \int_0^\frac{\pi}{4} \frac{\sin^2 x \cos^2 x}{\cos^6 x \left( \tan^3 x + 1 \right)^2}dx\]
\[ = \int_0^\frac{\pi}{4} \frac{\tan^2 x \sec^2 x}{\left( \tan^3 x + 1 \right)^2}dx\]

Put

\[\tan^3 x + 1 = z\]

\[\therefore 3 \tan^2 x \sec^2 xdx = dz\]
\[ \Rightarrow \tan^2 x \sec^2 xdx = \frac{dz}{3}\]

When

\[x \to 0, z \to 1\]

When

\[x \to \frac{\pi}{4}, z \to 2\]

\[\therefore I = \frac{1}{3} \int_1^2 \frac{dz}{z^2}\]
\[ = \left.\frac{1}{3} \times - \frac{1}{z}\right|_1^2 \]
\[ = - \frac{1}{3}\left( \frac{1}{2} - 1 \right)\]
\[ = - \frac{1}{3} \times \left( - \frac{1}{2} \right)\]
\[ = \frac{1}{6}\]