Evaluate the following as true or false.

$$\int \mathrm{tan}(2x)dx=-\frac{1}{2}\mathrm{ln}|\mathrm{cos}(2x)|+C$$

$$\int \mathrm{tan}(2x)dx=-\frac{1}{2}\mathrm{ln}|\mathrm{cos}(2x)|+C$$

Harper George
2022-09-05
Answered

Evaluate the following as true or false.

$$\int \mathrm{tan}(2x)dx=-\frac{1}{2}\mathrm{ln}|\mathrm{cos}(2x)|+C$$

$$\int \mathrm{tan}(2x)dx=-\frac{1}{2}\mathrm{ln}|\mathrm{cos}(2x)|+C$$

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Emilia Boyle

Answered 2022-09-06
Author has **10** answers

Solution: True

$$\int \mathrm{tan}(2x)dx=-\frac{1}{2}\mathrm{ln}|\mathrm{cos}2x|+C$$

$$I=\int \mathrm{tan}(2x)dx$$, substitute 2x=t, 2dx=dt

$$I=\frac{1}{2}\int \mathrm{tan}(t)dt$$

$$I=\frac{1}{2}\mathrm{ln}|\mathrm{sec}x|+c,\mathrm{sec}x=\frac{1}{\mathrm{cos}x}=(\mathrm{cos}x{)}^{-1},\mathrm{ln}{a}^{m}=m\mathrm{ln}a$$

$$\therefore I=-\frac{1}{2}\mathrm{ln}|\mathrm{cos}x|+C$$

$$\int \mathrm{tan}(2x)dx=-\frac{1}{2}\mathrm{ln}|\mathrm{cos}2x|+C$$

$$I=\int \mathrm{tan}(2x)dx$$, substitute 2x=t, 2dx=dt

$$I=\frac{1}{2}\int \mathrm{tan}(t)dt$$

$$I=\frac{1}{2}\mathrm{ln}|\mathrm{sec}x|+c,\mathrm{sec}x=\frac{1}{\mathrm{cos}x}=(\mathrm{cos}x{)}^{-1},\mathrm{ln}{a}^{m}=m\mathrm{ln}a$$

$$\therefore I=-\frac{1}{2}\mathrm{ln}|\mathrm{cos}x|+C$$

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