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Question Number 21708 by Isse last updated on 01/Oct/17

$${\int }_0^{0.5}2{tan}^{2}2tdt$$

Answered by $@ty@m last updated on 01/Oct/17

$${\int }_0^{0.5}2(\sec {}^{2}2t-1)dt$$$${\int }_0^{0.5}2\sec {}^{2}2tdt-{\int }_0^{0.5}2dt$$$${\left[2\frac{\tan 2t}{2}\right]}_0^{0.5}-2{\left[t\right]}_0^{0.5}$$$$\tan 1-\tan 0-2\times 0.5$$$$\tan 1-1$$

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