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Question Number 52233 by de best last updated on 04/Jan/19

$$pleasecanyouhelpmewiththis$$$$caculus:\int \frac{1}{{cos}^{2}x}dx$$

Answered by MJS last updated on 04/Jan/19

$$\mathrm{this}\mathrm{is}\mathrm{a}\mathrm{standard}\mathrm{integral}$$$$f\left(x\right)=\tan x$$$$f^{\prime }\left(x\right)=\frac{1}{{\cos }^{2}x}$$$$\Rightarrow \int \frac{dx}{{\cos }^{2}x}=\tan x+C$$

Answered by peter frank last updated on 04/Jan/19

$$\int {sec}^{2}xdx=\tan x+G$$

Answered by peter frank last updated on 04/Jan/19

$$\int {sec}^{2}xdx=\tan x+G$$

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