Show that \(\frac{\mathrm{d}}{\mathrm{d}\theta} \tan^3 \theta\) may be expressed as \(3\tan^4 \theta + 3\sec^2 \theta - 3.\)
Hence or otherwise, find \(\displaystyle\displaystyle \displaystyle\int_{0}^{\frac{\pi}{4}} \tan^4 \theta \ \mathrm{d} \theta.\)


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