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$\\ Let \quad I=\int \tan ^{3} 2 x \cdot \sec 2 x d x$ $=\int \tan ^{2} 2 x \cdot \sec 2 x \cdot \tan 2 x d x$ \\ $\Rightarrow I=\int\left(\sec ^{2} 2 x-1\right) \sec 2 x \cdot \tan 2 x d x$ \\ Put sec $2 x=t$ \\ $2 \sec 2 x \cdot \tan 2 x d x=d t$ \\ Now sec $2 x .$ tan $2 x d x=\frac{d t}{2}$ \\ $I=\frac{1}{2} \int\left(t^{2}-1\right) d t$ $=\frac{1}{2}\left[\frac{t^{3}}{3}-t\right]+C$ $=\frac{1}{2}\left[\frac{\sec ^{3} 2 x}{3}-\sec 2 x\right]+C$$

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