Question Number 21708 by Isse last updated on 01/Oct/17
$$\int_{\mathrm{0}} ^{\mathrm{0}.\mathrm{5}} \mathrm{2}{tan}^{\mathrm{2}} \mathrm{2}{tdt} \\ $$
Answered by $@ty@m last updated on 01/Oct/17
![∫_0 ^(0.5) 2(sec^2 2t−1)dt ∫_0 ^(0.5) 2sec^2 2tdt−∫_0 ^(0.5) 2dt [2((tan2t )/2)]_0 ^(0.5) −2[t]_0 ^(0.5) tan1−tan0−2×0.5 tan1−1](Q21710.png)
$$\int_{\mathrm{0}} ^{\mathrm{0}.\mathrm{5}} \mathrm{2}\left(\mathrm{sec}\:^{\mathrm{2}} \mathrm{2}{t}−\mathrm{1}\right){dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{0}.\mathrm{5}} \mathrm{2sec}\:^{\mathrm{2}} \mathrm{2}{tdt}−\int_{\mathrm{0}} ^{\mathrm{0}.\mathrm{5}} \mathrm{2}{dt} \\ $$$$\left[\mathrm{2}\frac{\mathrm{tan2}{t}\:}{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{0}.\mathrm{5}} −\mathrm{2}\left[{t}\right]_{\mathrm{0}} ^{\mathrm{0}.\mathrm{5}} \\ $$$$\mathrm{tan1}−\mathrm{tan0}−\mathrm{2}×\mathrm{0}.\mathrm{5}\:\: \\ $$$$\mathrm{tan1}−\mathrm{1}\: \\ $$